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Thank You! The intent was to make sure that people think twice before applying any standard integration technique. And yep, it’s good for the brain. 🙂

ReplyBy Math Vault | Calculus

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- Integration Series: The Overshooting Method

Hey! Greeting from a bunch of *post-April-Fool bunnies* who never managed to get their modules published on time — because they were simply too busy eating *Easter eggs* and *shooting for the moon*. 🙂

Moving onto a more serious topic though, if you’re currently into (or have been into) this thing called **integral calculus**, you might know from first-hand experience that integration in general is no easy task. After all, most of us spend at least a semester honing different methods of integration, which include — among others — the techniques of **Substitution**, **Partial Fraction** and I**ntegration By Parts**.

In what follows though, we will share with you a powerful technique for finding **antiderivatives** — possibly without paper and pen. We call this technique the **Overshooting Method**, which — as we shall see later — can even be just as efficient as several **standard integration techniques** combined.

So what is this *marvelous* technique you ask? Well, the short answer is that it can be found in the header image above. What? **Overshooting**, of course! 🙂

More specifically, given a function $f$, the **Overshooting Method** consists in finding an antiderivative of $f$ first by *guessing* a **potential candidate**, and then checking to see how *close* the candidate *differentiates* to $f$.

In the event where the derivative is off, but *only* by an **additional term** or a **multiple**, then additional steps can be implemented to correct this discrepancy, thereby transforming a potential candidate into a *valid* antiderivative of $f$.

All right. That’s about as intuitive and accurate as it gets. In **formal terms** though, this is what the technique would look like *in a nutshell*:

Theorem 1 — The Overshooting Method

Given a function $f$ defined on an interval $I$, if there exists another function $F^*$ such that:

- $(F^*)’ = kf$ on $I$ for some
*non-zero*number $k$ (i.e., $F^*$ is off by a**multiple**), then the function $\dfrac{F^*}{k}$ constitutes a valid antiderivative of $f$ on $I$. - $(F^*)’=f + g$ on $I$ for some function $g$, with $\displaystyle \int g$ being one of its own antiderivatives on $I$ (i.e., $F^*$ is off by a
**term**that is itself*antidifferentiable*), then the function $\displaystyle F^* – \int g$ constitutes a valid antiderivative of $f$ on $I$.

OK. Enough of the mathematical jargon? Let’s move on to how we can apply the Overshooting Method to integrate all kinds of functions then!

As it turns out, when a potential antiderivative is off by *only* a **multiple**, it’s not hard to readjust it so as to produce a valid antiderivative. In fact, learning to adjust for a multiple paves the way for more complex applications of the Overshooting Method, which can take a bit more* insight* and *ingenuity* to be executed elegantly.

With that in mind, here are *16 examples* illustrating how the Overshooting Method allows for certain **trigonometric.** **exponential, power functions**.and **trig-substitution functions** to be integrated with ease. 🙂

To find an antiderivative of a function like $6\cos (2x+4)$, for instance, we start by noticing that since this function behaves very much like $\cos x$, we can try to integrate it *as if* we were dealing with $\cos x$. With this insight, the idea of $\sin (2x+4)$ being a *potential* antiderivative of $6\cos (2x+4)$ naturally comes to mind.

Before we move on though, let’s check how well this idea works out:

\begin{align*}[\sin (2x+4)]’ & =\cos (2x+4)(2x+4)’ \\ & = 2\cos (2x+4)= \frac{6\cos (2x+4)}{3} \end{align*}

So, $\sin (2x+4)$ is *off* by a multiple of $\dfrac{1}{3}$ as a result of chain rule. To get rid this multiple, we multiply our *potential antiderivative* by $3$, yielding that:

\begin{align*}\displaystyle [3\sin (2x+4)]’ & =3\, [\sin (2x+4)]’\\ & = 3\, \frac{6\cos (2x+4)}{3} \\ & = 6\cos (2x+4) \qquad (\text{for all } x \in \mathbb{R}) \end{align*}

Hence, $\displaystyle \int 6\cos (2x+4)\, dx = 3 \sin (2x+4) + C$ for all $x \in \mathbb{R}$ — as desired.

How about a function such as $\pi \sin (10x)$? Well, we can always begin by pulling the *coefficient* $\pi$ outside the integral:

$$ \int \pi \sin (10x)\, dx = \pi \int \sin (10x)\, dx$$

Presumably, since $\sin (10x)$ behaves very much like $\sin x$, the first potential antiderivative that comes to mind would be $-\cos (10x)$. To test it, we need to see what it *actually* differentiates to:

$$ [- \cos (10x)]’ = \sin (10x) (10x)’ = 10 \sin(10x) $$

This means that our candidate is off by a multiple of $10$. Dividing it however by $10$ yields that:

\begin{align*} \left(\frac{- \cos (10x)}{10}\right)’ & = \frac{1}{10}\, [- \cos (10x)]’ \\ &= \frac{1}{10} \, [10 \sin(10x)] \\ & = \sin (10x) \qquad (\text{for all }x \in \mathbb{R})\end{align*}

which is exactly what we needed. Putting everything together, we get that $\displaystyle \int \pi \sin (10x)\, dx = \pi \int \sin (10x)\, dx $ $\displaystyle = \pi \, \frac{- \cos (10x)}{10} + C$ for all $x \in \mathbb{R}$.

For functions such as $\displaystyle \dfrac{1}{2} \sec^2 (4-x)$, we begin by pulling out the *extraneous* coefficent:

$$ \int \dfrac{1}{2} \sec^2 (4-x) \, dx= \dfrac{1}{2} \int \sec^2 (4-x) \, dx $$

Here, since $\sec^2 (4-x)$ behaves very much like $\sec^2 x$, we choose $\tan (4-x)$ as a potential antiderivative to start with. To be sure, we actually have that:

$$ [\tan (4-x)]’ = – \sec^2 (4-x) $$

which means that we are off by a *negative sign*. In this case, simply *negating* our candidate function would do the trick:

$$ [-\tan (4-x)]’ = \sec^2 (4-x) $$

Therefore,

\begin{align*} \int \dfrac{1}{2} \sec^2 (4-x) \, dx & = \dfrac{1}{2} \int \sec^2 (4-x) \, dx \\ & = \frac{1}{2}\, [-\tan (4-x)] + C\end{align*}

(**Bonus**: Can you figure out what is the *largest* domain under which this equality holds? 🙂 )

For functions such as $3 \sec (2x+ \pi) \tan (2x+ \pi)$, we begin by noticing that since this falls into the family of $\sec x \tan x$, a potential antidervative that comes to mind would be $\sec (2x+ \pi)$. To be sure, we have that:

\begin{align*} [\sec (2x + \pi)]’ & = \sec (2x+\pi) \tan (2x+\pi) (2x+\pi)’ \\ & = 2 \sec (2x+\pi) \tan (2x+\pi) \end{align*}

So almost there, except that we need to have $3$ as our **leading coefficient** instead of $2$, and a bit of reflection shows that multiplying our candidate by $\dfrac{3}{2}$ would do:

\begin{align*} \left( \frac{3}{2}\sec (2x + \pi) \right)’ & = \frac{3}{2} \, 2 \sec (2x+\pi) \tan (2x+\pi) \\ & = 3 \sec (2x+\pi) \tan (2x+\pi) \end{align*}

And that’s a *homerun*! Hence $\displaystyle \int 3 \sec (2x+ \pi) \tan (2x+ \pi) \, dx =$ $\displaystyle \frac{3}{2}\sec (2x + \pi) + C$ (where $2x + \pi \ne \frac{\pi}{2} + k\pi$ for some *integer* $k$).

For functions like $-6 \csc^2 (5x-4)$, we start by noticing that since it pretty much behaves like $\csc^2 x$ (which antidifferentiates to $-\cot x$ — by the way), the idea of $-\cot (5x-4)$ being one of its antiderivatives becomes more than a *remote reality*. However, we need to verify this idea in practice:

$$[-\cot (5x-4)]’= \csc^2 (5x-4) (5x-4)’ = 5 \csc^2 (5x-4) $$

To correct the leading coefficient from $5$ to $-6$, we multiply our guess by $\dfrac{-6}{5}$, yielding that:

\begin{align*} \left( \frac{-6}{5} [- \cot (5x-4)] \right)’ & = \frac{-6}{5} \, [5 \csc^2 (5x-4)] \\ & = -6 \csc^2 (5x-4)\end{align*}

Done! Hence $\int -6 \csc^2 (5x-4) \, dx =$ $\frac{-6}{5}\, [- \cot (5x-4)] + C =$ $\frac{6}{5} \cot (5x-4) +C\,$ (where $5x-4 \ne k \pi$ for some integer $k$).

OK. What about $\cot (2x-e)$? Let’s see… looking at the **table of integrals**, we see that $\cot x$ integrates to $\ln |\sin x|$, which suggests that $\ln |\sin (2x-e)|$ might work out very well as a candidate. Let’s check what it differentiates to:

$$ \left[\ln |\sin (2x-e)|\right]’ = \frac{\cos(2x-e)}{\sin (2x-e)}(2x-e)’= 2\cot (2x-e)$$

Pretty close! Because in thise case, dividing both sides of the equation by $2$ would do:

$$ \left[\frac{1}{2}\ln |\sin (2x-e)| \right]’ = \cot (2x-e)$$

Therefore, $\displaystyle \int \cot (2x-e) \, dx = \frac{1}{2}\ln |\sin (2x-e)| + C$ (where $2x-e \ne k\pi$ for some integer $k$).

For functions such as $\displaystyle -5.6e^{-2x+4}$, once we notice its *similarity* with $\displaystyle e^x$, trying out $\displaystyle e^{-2x+4}$ becomes our *first line of attack*:

$$(e^{-2x+4})’=e^{-2x+4} \, (-2x+4)’ = -2 e^{-2x+4} $$

Now, multiplying both sides by $\dfrac{5.6}{2}$ yields:

$$\left( \frac{5.6}{2} \, e^{-2x+4} \right)’=\frac{5.6}{2} (-2 e^{-2x+4}) = -5.6 e^{-2x+4}$$

And that’s it! Hence $\displaystyle \int -5.6e^{-2x+4} \ dx = \frac{5.6}{2}(e^{-2x+4}) + C$ (for all $x \in \mathbb{R}$).

So that was for the **natural base** $e$. For exponential functions of other bases like $34 \pi^{3x-1}$, we can always start by taking out the *annoying* leading coefficient:

$$ \int 34 \pi^{3x-1} \, dx = 34 \int \pi^{3x-1} \, dx$$

Now, what function could possibly *differentiate* to $\displaystyle \pi^{3x-1}$ ? Out of curiosity, let’s just try the function *itself*:

$$\left(\pi^{3x-1}\right)’ = \pi^{3x-1} \ln \pi \, (3x-1)’ = (3 \ln \pi) \pi^{3x-1}$$

In which case, all that’s left to do is to *divide both sides* by $3 \ln \pi$:

$$\left(\frac{\pi^{3x-1}}{3 \ln \pi}\right)’ = \pi^{3x-1}$$

yielding that $\displaystyle \int 34 \pi^{3x-1} \, dx =$ $\displaystyle 34 \int \pi^{3x-1} \, dx =$ $\displaystyle 34 \, \frac{\pi^{3x-1}}{3 \ln \pi} +C\,$ (for all $x \in \mathbb{R}$).

Here’s another one for you: $e\sqrt{7x-9}$. *Strange* huh?

Actually, by inspection, this really just looks like $\sqrt{x}$, so let’s try out $\displaystyle \frac{(7x-9)^{\frac{3}{2}} }{\frac{3}{2}}$:

$$ \left( \frac{(7x-9)^{\frac{3}{2}} }{\frac{3}{2}} \right)’= \sqrt{7x-9}\, (7x-9)’ = 7 \sqrt{7x-9}$$

Here, we need to turn the $7$ into a $\displaystyle e$, and a bit of thought reveals that multiply both sides by $\dfrac{e}{7}$ would do the trick:

$$\left( \frac{e}{7}\, \frac{(7x-9)^{\frac{3}{2}} }{\frac{3}{2}} \right)’= \frac{e}{7} \, 7 \sqrt{7x-9} = e \sqrt{7x-9}$$

All good! Hence $\displaystyle \int e \sqrt{7x-9} \, dx = \frac{e}{7} \frac{(7x-9)^{\frac{3}{2}} }{\frac{3}{2}} + C$ $\displaystyle = \frac{2e}{21}(7x-9)^{\frac{3}{2}} + C$ (where $7x-9>0$).

With the square root now *conquered*, let’s turn it upside down (literally) and look at a function like $\displaystyle \frac{3\pi}{ \sqrt{-5.5x+3}}$ instead. First, pulling out the extraneous *numerator* yields:

$$\int \frac{3\pi} { \sqrt{-5.5x+3}} \, dx = 3\pi \int (-5.5x+3)^{-\frac{1}{2}} \, dx$$

Here, we see that we’re dealing with something similar to $\displaystyle x^{-\frac{1}{2}}$ — which integrates to $\displaystyle \frac{x^{\frac{1}{2}}}{\frac{1}{2}}$. This prompts us to try out $\displaystyle \frac{(-5.5x+3)^{\frac{1}{2}}}{\frac{1}{2}}$ as a *potential antiderivative*:

$$ \left( \frac{(-5.5x+3)^{\frac{1}{2}}}{\frac{1}{2}} \right)’ = -5.5 (-5.5x+3)^{-\frac{1}{2}}$$

Dividing on both sides by $-5.5$ then yields:

$$ \left( \frac{1}{-5.5}\frac{(-5.5x+3)^{\frac{1}{2}}}{\frac{1}{2}} \right)’ = (-5.5x+3)^{-\frac{1}{2}}$$

All good! Putting everything together, we have that:

\begin{align*} \int \frac{3\pi}{ \sqrt{-5.5x+3}} \, dx & = 3\pi \int (-5.5x+3)^{-\frac{1}{2}} \, dx \\ & = 3\pi \frac{1}{-5.5}\frac{(-5.5x+3)^{\frac{1}{2}}}{\frac{1}{2}} + C \\ & = \frac{6 \pi}{-5.5} (-5.5x+3)^{\frac{1}{2}} + C \qquad (\text{for}-5.5x+3>0) \end{align*}

For functions such as $\displaystyle \frac{4}{17}(3x-5)^7$, instead of *expanding the terms*, we can try to integrate it *on the spot* by first pulling out the leading coefficient:

$$\int \frac{4}{17}(3x-5)^7 \, dx = \frac{4}{17} \int (3x-5)^7 \, dx$$

Here, we can integrate the function as if it were $\displaystyle x^7$, so $\displaystyle \frac{(3x-5)^8}{8}$ becomes a natural choice:

$$ \left( \frac{(3x-5)^8}{8} \right)’ =3\, (3x-5)^7$$

Dividing both sides of the equation by $3$ yields:

$$ \left( \frac{1}{3} \frac{(3x-5)^8}{8} \right)’ =(3x-5)^7$$

Therefore,

\begin{align*} \int \frac{4}{17}(3x-5)^7 \, dx & = \frac{4}{17} \int (3x-5)^7 \, dx \\ & = \frac{4}{17}\frac{1}{3} \frac{(3x-5)^8}{8}+C \\ & = \frac{(3x-5)^8}{102} + C \qquad \text{(for all $x \in \mathbb{R}$)} \end{align*}

OK. But what if the polynomial is found in the *denominator*, like $\displaystyle \frac{25}{(\pi x – e)^3}$? Oh well, let’s start by pulling out the *numerator* and see where we can go from there:

$$ \int \frac{25}{(\pi x – e)^3} \, dx =25 \int (\pi x-e)^{-3} \, dx$$

By inspection, that function looks a bit similar to $x^{-3}$, so $\displaystyle \frac{(\pi x-e)^{-2}}{-2}$ seems like a good way to go:

$$\left( \frac{(\pi x-e)^{-2}}{-2} \right)’ = \pi (\pi x – e)^{-3} $$

Dividing both sides by $\pi$, we get:

$$\left( \frac{(\pi x-e)^{-2}}{-2\pi} \right)’ = (\pi x – e)^{-3} $$

which is what we expected. Putting everything together, we conclude that $\int \frac{25}{(\pi x – e)^3}\, dx = 25 \int (\pi x-e)^{-3} \, dx = 25 \, \frac{(\pi x-e)^{-2}}{-2\pi} +C$ (where $\pi x – e \ne 0$).

With a *plain old* reciprocal function like $\displaystyle \frac{2}{100-x}$, we can integrate as if it were $\displaystyle \frac{1}{x}$. This would prompt $\ln |100-x|$ as a potential candidate:

$$ \left( \ln|100-x| \right)’ = \frac{-1}{100-x}$$

To correct the *numerator* from $-1$ to $2$, we multiply both sides of the equation by $-2$, yielding that:

$$ \left( -2 \ln|100-x| \right)’ = \frac{2}{100-x}$$

And that’s it! $\displaystyle \int \frac{2}{100-x} \, dx = -2 \ln|100-x| + C$ (where $100-x \ne 0$).

Let’s move on to some function we haven’t quite seen yet: $\displaystyle \frac{25}{\sqrt{1-(3x)^2}}$. What to do? Well, let’s start by pulling out the *constant* as usual:

$$ \int \frac{25}{\sqrt{1-(3x)^2}} \, dx = 25 \int \frac{1}{\sqrt{1-(3x)^2}} \, dx $$

and if that function looks strangely *familiar*, it’s because it belongs to the same family as the function $\displaystyle \frac{1}{\sqrt{1-x^2}}$ — which incidentally integrates to $\arcsin x$. This tells us that $\arcsin (3x)$ could be a good candidate for the antiderivative, but then, we need to test it:

$$ \left[ \arcsin (3x) \right]’ = \frac{3}{\sqrt{1-(3x)^2}}$$

That’s pretty close! Further dividing both sides by $3$ yields:

$$ \left( \frac{\arcsin (3x)}{3} \right)’ = \frac{1}{\sqrt{1-(3x)^2}}$$

which means that $ \displaystyle \int \frac{25}{\sqrt{1-(3x)^2}} \, dx =$ $\displaystyle 25 \int \frac{1}{\sqrt{1-(3x)^2}} \, dx$ $\displaystyle = 25 \, \frac{\arcsin (3x)}{3} + C$ (where $-1<3x<1$).

Here’s a neater one: $\displaystyle \frac{5}{1+(4x)^2}$. To start, pulling out the *numerator* yields:

$$ \int \frac{5}{1+(4x)^2} \, dx = 5 \int \frac{1}{1+(4x)^2} \, dx$$

which produces a function reminiscent of $\displaystyle \frac{1}{1+x^2}$. And since that last one integrates to $\arctan x$, choosing $\arctan (4x)$ as a *potential antiderivative* seems like a good way to go:

$$ \left[ \arctan (4x) \right]’ = \frac{4}{1+(4x)^2}$$

Once here, dividing both sides by $4$ yields:

$$ \left[ \frac{\arctan (4x)}{4} \right]’ = \frac{1}{1+(4x)^2}$$

which means that $\displaystyle \int \frac{5}{1+(4x)^2} \, dx =$ $\displaystyle 5 \int \frac{1}{1+(4x)^2} \, dx$ $\displaystyle = 5 \, \frac{\arctan (4x)}{4} +C$ (for all $x \in \mathbb{R}$).

All right. For those looking for some *gibberish-looking* integral, this is it!

$$ \frac{3e}{|x|\sqrt{(2x)^2 – 1}}$$

Clueless, let’s start by pulling out the *numerator* as usual first:

$$ \int \frac{3e}{|x|\sqrt{(2x)^2 – 1}} \,dx = 3e \int \frac{1}{|x|\sqrt{(2x)^2 – 1}} \, dx$$

Looking at the **table of integrals**, we notice that this is similar to the function $ \displaystyle \frac{1}{|x|\sqrt{x^2 – 1}}$ — which integrates to $\text{arcsec} x$. This suggests that $\text{arcsec} (2x)$ might just pull it off:

\begin{align*} \left[ \text{arcsec} (2x) \right]’ & = \left[ \frac{1}{|\Box|\sqrt{(\Box)^2-1}}\right]_{\Box = 2x} (2x)’ \\ & = \frac{2}{|2x|\sqrt{(2x)^2-1}} \\ & =\frac{1}{|x|\sqrt{(2x)^2-1}} \end{align*}

In other words,

\begin{align*}\int \frac{3e}{|x|\sqrt{(2x)^2 – 1}} \,dx & = 3e \int \frac{1}{|x|\sqrt{(2x)^2 – 1}} \, dx \\ & = 3e \, \text{arcsec} (2x) + C \qquad \text{(for $2x \in (-\infty, -1) \cup (1, \infty)$)} \end{align*}

So there you have it! That concludes the treatment of the Overshooting Method on **adjusting for multiples**. 🙂

Of course. The Overshooting Method does not end here, for as mentioned earlier, if a function *suspected* to be an antiderivative is off by *only* an **additional term** — being it a *constant* or *otherwise* — then adjustments can still be made to eliminate the unwanted term, thereby turning a potential antiderivative into an *actual* antiderivative. Indeed, this form of Overshooting works particularly well with functions such as:

**Polynomial-Exponentials**: Products of*polynomial*and*exponential functions*, including those that form the basis of**Gamma Function**and**Gamma distribution**.**Polynomial-Logarithms**: Products of*polynomial*and*logarithmic functions*— to be distinguished from**logarithmic polynomials**.**Polynomial-Trigonometrics**: Products of*polynomial*and*trigonometric functions*— not to be confused with**trigonometric polynomials**.

Intrigued? All right. Let’s get on to it!

Ever wonder how the canonical antiderivative of $\ln x$ is proved? Well here’s is a good one for you, because it essentially boils down to finding a function that has the *exquisite* property of *differentiating* to $\ln x$.

By inspection, $\ln x$ itself definitely won’t do, as $ \displaystyle (\ln x)’=\frac{1}{x}$. In contrast, with $x \ln x$, there’s still a *glimpse of hope*. This is because by the **Product Rule for Derivatives**:

$$(x\ln x)’ = \ln x + x \, \frac{1}{x} = \ln x +1$$

Goodness! See what just happened there? $ x \ln x$ was indeed quite spot on, but still slightly off by a constant of $1$ though. To get rid of this unwanted $1$, we need to readjust $x\ln x$ so that it differentiates to $\ln + 1 – 1$ instead. A bit of reflection then shows that subtracting an $x$ from $x \ln x$ could be a way to go:

$$ (x \ln x – x)’ = \ln x + 1 – 1 = \ln x $$

And now we are talking! For we’ve just showed that $\displaystyle \int \ln x \, dx = x \ln x – x + C$ (for $x>0$). The next time you spot this line in a **table of integrals**, you should be able to *derive* (excuse the pun) a whole new meaning from it. Either way, the general idea doesn’t change:

When you hack the Product Rule, you get the Overshooting Method for integrals.

To add an extra layer of complexity, let’s consider the function $x^5 \ln (4x)$ instead. Let’s see… are we out of luck this time maybe?

Not really. Exploiting the **Product Rule** as always, we suspect that $\displaystyle \frac{x^6}{6} \ln (4x)$ might work out just fine:

\[ \left( \frac{x^6}{6} \ln (4x) \right)’ = x^5 \ln (4x) + \frac{x^6}{6} \frac{4}{4x} = x^5 \ln (4x) + \frac{x^5}{6} \]

Hmm, that’s not too bad, but there’s an *unwanted term* popping out, and it’s not a constant either. Fortunately though, we can still try to adjust our candidate a bit, so that it *differentiates* to $\displaystyle x^5 \ln (4x) + \frac{x^5}{6} – \frac{x^5}{6}$. After a bit of thought and experimentation, we conclude that *subtracting* the candidate by *an antiderivative of* $ \displaystyle \frac{x^5}{6}$ — say $\displaystyle \frac{x^6}{36}$ — should take care of this:

$$ \left( \frac{x^6}{6} \ln (4x) – \frac{x^6}{36} \right)’ = x^5 \ln (4x) + \frac{x^5}{6} – \frac{x^5}{6} = x^5 \ln (4x) $$

Still alive and kicking! We’ve just shown that $ \displaystyle \int x^5 \ln (4x) \, dx = \frac{x^6}{6} \ln (4x) – \frac{x^6}{36} + C$ (where $4x>0$).

Now that we get the *gist* of the technique, we can proceed to *overshoot* integrals on the top of our head — and doing so at a *faster speed* even. Consider a function like $xe^x$ for example: What could a function *differentiating to it* possibly be?

Let’s see… exploiting the **Product Rule** and the fact that $e^x$ differentiates to itself, we suspect that the function $xe^x$ *itself* might be worthy of trying:

$$ (xe^x)’ = xe^x + e^x $$

That’s pretty good! To eliminate the unwanted $e^x$, we try subtracting our candidate by $e^x$, yielding that:

$$ (xe^x – e^x)’ = xe^x + e^x – e^x = xe^x $$

And that’s perfection! So we therefore conclude that $\displaystyle \int xe^x \, dx =$ $\displaystyle xe^x – e^x + C$ (for all $x \in \mathbb{R}$).

For functions such as $x \cos x$, we start by *guessing* a function that can potentially differentiate to it. Using the fact that $\cos x$ integrates to $\sin x$, we suspect that trying out $x \sin x$ could be a viable step forward:

$$ [x \sin x]’ = x \cos x + \sin x$$

To make the $\sin x$ disappear, we subtract our candidate by an *antiderivative* of $\sin x$, say, $-\cos x$:

$$ [x \sin x – (-\cos x)]’ = x \cos x + \sin x – \sin x = x \cos x$$

And mystery solved! With $\displaystyle \int x \cos x \, dx =$ $\displaystyle x \sin x +\cos x + C$ (for all $x \in \mathbb{R}$). Neat!

Taking from what we have just learnt, let’s try to see if we can spot a *pattern* that generalizes our previous findings a bit: suppose that we have two functions $f$ and $g$, then is there a *systematic* way of finding an antiderivative of $fg$ — *in general*?

Surprisingly, the answer is *no*. 🙁 However, in the case where $g$ is *differentiable*, and $f$ is *antidifferentiable* (with $F$ being one of its antiderivatives), then, exploiting the **Product Rule** again, we see that the function $Fg$ differentiates to something *close* to $fg$:

$$ (Fg)’ = fg + Fg’ $$

which means that *if* $Fg’$ is itself *antidifferentiable* (with $\int Fg’$ being one of its antiderivatives), then the **Overshooting Method** provides a way of eliminating this *unwanted* term with a *not-to-shabby* adjustment:

$$\left( Fg\; – \int Fg’ \right)’= fg + Fg’ – Fg’ = fg$$

Or alternatively:

$$ \int fg = Fg \; – \int Fg’ + C $$

And *that*, is our rendition of the *(in)famous* formula for Integration By Parts. 🙂

For example, if we were to integrate the **polynomial-trigonometric function** $x\sin x$, we can start by *splitting* it into $\sin x$ and $x$. From there, *antidifferentiating* the *first* followed by *differentiating* the *second*, we get that:

\begin{align*} \int x \sin x \, dx & = x (- \cos x)\, – \int – \cos x \, dx \\ & = \; – x \cos x \, +\, \sin x \, + \, C \qquad (\text{for all }x \in \mathbb{R}) \end{align*}

which is pretty neat. No more $u \ du \ v \ dv$ nonsense! And to add a bit more to the *goodness*, since a **quotient** is really a disguised form of a *product*, this suggests that in *some* circumstances, we can also integrate a quotient using this **Integration By Parts** formula — by splitting it into a *product* of two terms.

For example, to integrate $\displaystyle \frac{\ln x}{x}$, we begin by breaking it into $\displaystyle \frac{1}{x}$ and $\ln x$. Once here, *antidifferentiating* the *first* followed by *differentiating* the *second* yields that:

$$ \int \frac{\ln x}{x} \, dx = (\ln x)^2 – \int \frac{\ln x}{x} \, dx$$

What? Moving in *circle*? Not really, for if we add $\displaystyle \int \frac{\ln x}{x} \, dx$ on *both* sides of the equation, *dividing by *$2$ would then yield:

$$ \int \frac{\ln x}{x} \, dx = \frac{(\ln x)^2}{2} + C \qquad (\text{for } x>0)$$

Yeah. We know. It’s crazy, but there is a even *better* way, for had we tried to *overshoot* this integral with $(\ln x)^2$ as our *first guess*, we would have finished the integration *in a few seconds*!

And with all that goodness, we can now conclude the section with a *formal* summary of our formula on **Integration By Parts**:

Theorem 2 — Integration By Parts

Given two functions $f(x)$ and $g(x)$ defined on an interval $I$, if the following *three* conditions are met:

- $f$ is
*antidifferentiable (*with $F(x)$ being one of its antiderivatives on $I$). - $g$ is
*differentiable (*with $g'(x)$ denoting its derivative on $I$). - The function resulting from
*antidifferentiating $f$ and differentiating $g$*is*itself*antidifferentiable (with $\displaystyle \int F(x) g'(x)$ denoting one of its antiderivatives on $I$).

then $\displaystyle \int f(x)g(x) \, dx = F(x) g(x) – \int F(x) g'(x) +C$ on $I$.

All right. Now that we have discovered how the **Overshooting Method** allows for adjusting for both **multiples** and **additional terms**, it’s about time to *combine* these two approaches together and — without much *handholding* — demonstrates its full potential in all its glory. 🙂

So without further ado, here are *nine* functions which can be integrated via Overshooting, but which require a bit of **algebraic manipulations** and **ingenuity** before they can be tackled appropriately.

\begin{align*} x \sqrt{2x – 3} & = \frac{1}{2} (2x) \sqrt{2x-3} \\ & = \frac{1}{2} (2x-3 + 3) (2x-3)^{\frac{1}{2}} \\ & = \frac{1}{2} \left[ (2x-3)^{\frac{3}{2}}+3(2x-3)^{\frac{1}{2}}\right] \\ & = \frac{1}{2} (2x-3)^{\frac{3}{2}} +\frac{3}{2} (2x-3)^{\frac{1}{2}} \end{align*}

which means that:

\begin{align*} \int x \sqrt{2x-3} \, dx & = \frac{1}{2} \int (2x-3)^{\frac{3}{2}} \, dx + \frac{3}{2} \int (2x-3)^{\frac{1}{2} } \, dx \\ & = \frac{1}{2} \left[ \frac{1}{2} \frac{ (2x-3)^{\frac{5}{2}} }{\frac{5}{2} }\right] + \frac{3}{2} \left[ \frac{1}{2} \frac{ (2x-3)^{\frac{3}{2}} }{\frac{3}{2}} \right] + C \\ & = \frac{1}{10} (2x-3)^{\frac{5}{2}} + \frac{1}{2} (2x-3)^{\frac{3}{2}} + C \qquad (\text{for } 2x-3>0) \end{align*}

\begin{align*} \frac{2x}{3x-1} & = \frac{2}{3}\, \frac{3x}{3x-1} = \frac{2}{3}\, \frac{3x -1 +1}{3x – 1} \\ &= \frac{2}{3} \left[ 1 + \frac{1}{3x -1} \right] \\& = \frac{2}{3} + \frac{2}{3}\, \frac{1}{3x-1} \end{align*}

Therefore,

\begin{align*} \int \frac{2x}{3x-1} \, dx & = \frac{2}{3} \int 1 \, dx + \frac{2}{3} \int \frac{1}{3x-1} \, dx \\ & = \frac{2}{3}x + \frac{2}{9} \ln |3x-1| + C \qquad (\text{for }3x-1 \ne 0)\end{align*}

\begin{align*} \frac{1}{(2x)^2 + 15} & = \frac{1}{15} \, \frac{1}{\frac{(2x)^2}{15} + 1} \\ & = \frac{1}{15} \, \frac{1}{(\frac{2}{\sqrt{15}}x)^2+1}\end{align*}

Hence,

\begin{align*} \int \frac{1}{(2x)^2 + 15} \, dx & = \frac{1}{15} \int \frac{1}{(\frac{2}{\sqrt{15}}x)^2+1} \, dx \\ & = \frac{1}{15} \frac{\arctan (\frac{2}{\sqrt{15}}x)}{\frac{2}{\sqrt{15}}} + C \\ & = \frac{1}{2\sqrt{15}} \arctan (\frac{2}{\sqrt{15}}x) + C \qquad (\text{for all }x \in \mathbb{R})\end{align*}

A bit of ingenuity shows that:

\begin{align*} [\sin (2x^2)]’ & = 4x \cos (2x^2) \qquad \Rightarrow \\ \left[\frac{\sin (2x^2)}{4} \right]’ & = x \cos (2x^2) \end{align*}

from which we can infer that:

\begin{align*} \int 6x \cos (2x^2) \, dx & = 6 \int x \cos (2x^2) \, dx \\ & = 6 \, \frac{\sin (2x^2)}{4} + C \\ & = \frac{3}{2} \sin (2x^2) + C \qquad (\text{for all }x \in \mathbb{R} )\end{align*}

\begin{align*} [(5-2x) \ln (5-2x)]’ & = – 2 \ln (5-2x) – 2 & \Rightarrow \\ \left[ (5-2x) \ln (5-2x) + 2x\right]’ & = -2 \ln (5-2x) & \Rightarrow \\ \left[ \frac{(5-2x) \ln (5-2x) + 2x}{-2} \right]’ & = \ln (5-2x) \end{align*}

In other words,

\begin{align*} \int \ln(5-2x) \, dx & = \frac{(5-2x) \ln (5-2x) + 2x}{-2} + C \\ & = \frac{-1}{2} (5-2x) \ln (5-2x) – x +C \qquad (\text{for }5-2x>0)\end{align*}

\begin{align*} ( x \arcsin x)’ & = \arcsin x + \frac{x}{\sqrt{1-x^2}} & \Rightarrow \\ \left( x \arcsin x – \frac{ (1-x^2)^{\frac{1}{2}}}{\frac{1}{2}} \right)’ & = \arcsin x + \frac{x}{\sqrt{1-x^2}} – (1-x^2)^{-\frac{1}{2}} (-2x) \\ & = \arcsin x + \frac{x}{\sqrt{1-x^2}} – {\color{red} \frac{-2x}{\sqrt{1-x^2}} } & \Rightarrow \\ \left( x \arcsin x + (1-x^2)^{\frac{1}{2}} \right)’ & = \arcsin x + \frac{x}{\sqrt{1-x^2}} – \frac{x}{\sqrt{1-x^2}} \\ & = \arcsin x \end{align*}

Hence $ \displaystyle \int \arcsin x \, dx = x \arcsin x + (1-x^2)^{\frac{1}{2}} + C \quad ( \text{where } -1<x<1)$.

Out of curiosity, we have that:

\begin{align*} (e^x \sin x)’ & = e^x \sin x + e^x \cos x \\ (e^x \cos x)’ & = – e^x \sin x + e^x \cos x \end{align*}

Therefore,

\begin{align*} (e^x \sin x + e^x \cos x)’ & = 2 e^x \cos x \qquad \Rightarrow \\ \left(\frac{e^x \sin x + e^x \cos x}{2}\right)’ & = e^x \cos x \end{align*}

Or equivalently,

\begin{align*} \int e^x \cos x \, dx = \frac{e^x \sin x + e^x \cos x}{2} + C \qquad (\text{for all }x \in \mathbb{R} )\end{align*}

Carry on with with what we left off by *combining* the derivatives of $\displaystyle e^x \sin x$ and $\displaystyle e^x \cos x$ in an *opposite* way, we have that:

\begin{align*} (e^x \sin x – e^x \cos x)’ & = 2 e^x \sin x \qquad \Rightarrow \\ \left(\frac{e^x \sin x – e^x \cos x}{2}\right)’ & = e^x \sin x \end{align*}

from which it follows that:

\begin{align*} \int e^x \sin x \, dx = \frac{e^x \sin x – e^x \cos x}{2} + C \qquad (\text{for all }x \in \mathbb{R} )\end{align*}

First, playing around with $\displaystyle \sec x$ and $\displaystyle \tan x$, we get that:

\begin{align*} (\sec x)’ & = \sec x \tan x\\ (\tan x)’ & = \sec^2 x & \Rightarrow \\ (\sec x + \tan x)’ & = \sec x \tan x + \sec^2 x \\ & = \sec x (\sec x + \tan x) \end{align*}

So the function $\displaystyle \sec x \tan x$ has the *peculiar* property of differentiating to $\sec x$ times *itself.* For cases like this, applying the **natural logarithm** to the function would kill the *unwanted* term:

\begin{align*} \left(\ln |\sec x + \tan x| \right)’ & = \frac{1}{\sec x + \tan x} \, (\sec x) (\sec x+\tan x) \\ & = \sec x\end{align*}

Done! Hence $\displaystyle \int \sec x \, dx = \ln |\sec x + \tan x| + C$ (where $\displaystyle x \ne \frac{\pi}{2}+ k \pi$ for some integer $k$).

And that’s a wrap!

So! As we can see, that was quite a range of integrals being solved there, and each one started *invariably* by making some innocent-looking** guesses**! Indeed, by learning the mechanics of the **Overshooting Method** — first on adjusting for **multiples**, and second on adjusting for **additional terms** — we manage to turn an *apparently-crude* method into an *indispensable tool* for finding antiderivatives of a plethora of functions ranging from **polynomial**, **exponential** to **logarithmic** and **trigonometric functions** (the list goes on!), and in so doing, we also established the **Integration By Parts** formula as a *special instance* of the Overshooting Method.

However, what’s perhaps more subtle is that Overshooting has the effect of *forcing *its user to develop insight about the **nature** of a function — and the different ways to navigate and maneuver around it. In effect, this kind of knowledge is of great importance as one advances through higher levels of mathematics, not to mention that it also prevents one from *blindly* applying the methods of **Substitution**, **Partial Fraction**, **Integration By Parts** and **Back Substitution** (or perhaps relying on a **table of integrals**) for functions which they don’t have a full and solid grasp on. As a surprisingly-efficient **heuristic** based primarily on **trial-and-error reasoning**, the Overshooting Method also has the advantage of sidestepping the conundrum of **switching between variables**, while at the same time allowing the **domain of antidifferentiability** to be mapped out more easily.

(of course, there are also other *indirect* benefits as a result of mastering the Overshooting Method, which includes — among others — the endowment of an *unfair advantage* in an integration bee 🙂 )

That’s being said, we also have to acknowledge the **limitations** of the Overshooting Method, in that it cannot solve for all integrals *once and for all*, as the effectiveness of the method depends heavily on the *quality* of the **initial guess**. While there are always functions whose antiderivatives cannot be expressed in simple terms *regardless* of the method being used, the traditional integration techniques are still very much relevant in integrating certain **specialized integrals** which cannot be solved easily by inspection — or for which a viable guess cannot be easily obtained without additional insight.

All right. Before we go, here’s an **interactive table** summarizing all the theories and examples covered thus far.

Given a function $f(x)$ defined on an interval $I$, if there exists another function $F^*(x)$ such that:

$$(F^*)'(x) = kf(x)$$

on $I$ for some *non-zero* number $k$, then:

$$\int f(x) \, dx =\frac{F^*}{k} + C \quad (\text{on } I ) $$

Given a function $f(x)$ defined on an interval $I$, if there exists another function $F^*(x)$ such that:

$$ (F^*)'(x)=f(x) + g (x) $$

on $I$ for some function $g(x)$ (with $\displaystyle \int g(x)$ being one of its own antiderivatives on $I$), then:

$$ \int f(x) \, dx = F^*(x) – \int g(x) + C \quad (\text{on } I )$$

Given two functions $f(x)$ and $g(x)$ defined on an interval $I$, if the following *three* conditions are met:

- $f$ is
*antidifferentiable*, with $F(x)$ being one of its antiderivatives on $I$. - $g$ is
*differentiable*, with $g'(x)$ denoting its derivative on $I$. - The function resulting from
*antidifferentiating $f$ and differentiating $g$*is*itself*antidifferentiable (with $\displaystyle \int F(x) g'(x)$ denoting one of its antiderivatives on $I$ ).

then $\displaystyle \int f(x)g(x) \, dx = F(x) g(x) – \int F(x) g'(x) +C$ on $I$.

**Cosine**: $\displaystyle 6\cos (2x+4)$**Sine**: $\displaystyle \pi \sin (10x)$**Secant**: $\displaystyle \frac{1}{2} \sec^2 (4-x)$

**Secant-Tangent**: $\displaystyle 3 \sec (2x+ \pi) \tan (2x+ \pi)$**Cosecant**: $\displaystyle -6 \csc^2 (5x-4)$**Cotangent**: $\displaystyle \cot (2x-e)$

**Natural Base**: $\displaystyle -5.6e^{-2x+4}$

**Arbitrary Base**: $\displaystyle 34 \pi^{3x-1}$

**Square Root**: $\displaystyle e\sqrt{7x-9}$**Reciprocal of Square Root**: $\displaystyle \frac{3\pi+2} { \sqrt{-5.5x+3}}$**Polynomial**: $\displaystyle \frac{4}{17}(3x-5)^7$

**Reciprocal of Polynomial**: $\displaystyle \frac{25}{(\pi x – e)^3}$**Plain Reciprocal**: $\displaystyle \frac{2}{100-x}$

**Type I**: $\displaystyle \frac{25}{\sqrt{1-(3x)^2}}$**Type II**: $\displaystyle \frac{5}{1+(4x)^2}$**Type III**: $ \displaystyle \frac{3e}{|x|\sqrt{(2x)^2 – 1}}$

**Logarithm**: $\displaystyle \ln x$**Polynomial-Logarithm**: $\displaystyle x^5 \ln (4x)$

**Polynomial-Exponential**: $\displaystyle xe^x$**Polynomial-Trigonometric**: $\displaystyle x \cos x$

**Polynomial-Trigonometric**: $\displaystyle x\sin x$**Logarithm over Polynomial**: $\displaystyle \frac{\ln x}{x}$

- $\displaystyle x \sqrt{2x – 3}$
- $\displaystyle \frac{2x}{3x-1}$
- $\displaystyle \frac{1}{(2x)^2 + 15}$
- $\displaystyle 6x \cos(2x^2)$
- $\displaystyle \ln (5-2x)$

- $\displaystyle \arcsin x$
- $\displaystyle e^x \cos x$
- $\displaystyle e^x \cos x$
- $\displaystyle \sec x$

And with that, let’s call it a day on overshooting, but then, we all know that the trip is not over yet, because now it’s your turn to *invent* some integrals to overshoot on your own!

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Math Vault April 7, 2016

Thank You! The intent was to make sure that people think twice before applying any standard integration technique. And yep, it’s good for the brain. 🙂

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