# Integration Series: The Overshooting Method

Calculus, College Math

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 Integration 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.

## The Overshooting Method — Basic Ideas

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!

## The Overshooting Method — Adjusting for Multiples

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 can be used to integrate certain elementary functions with ease. These include a good chunk of trigonometric functions, exponential functions, power functions and trig-substitution functions.

### Trigonometric Functions

#### Cosine Functions

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.

#### Sine Functions

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 $x \in \mathbb{R}$.

#### Secant Functions

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? 🙂 )

#### Secant-Tangent Functions

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 $k$).

#### Cosecant Functions

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 $k$).

#### Cotangent Functions

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 $k$).

### Exponential Functions

#### Natural Base

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 $x \in \mathbb{R}$).

#### Arbitrary Bases

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 $x \in \mathbb{R}$).

### Power Functions

#### Square Root Functions

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 $7x-9>0$).

#### Reciprocal of Square Root Functions

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*}

#### Polynomial Functions

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*}

#### Reciprocal of Polynomials

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 $\pi x – e \ne 0$).

#### Reciprocal Functions

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 $100-x \ne 0$).

### Trigonometric-Substitution Functions

#### Trig-Substitution Functions — Type I

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 $-1<3x<1$).

#### Trig-Substitution Functions — Type II

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 $x \in \mathbb{R}$).

#### Trig-Substitution Functions — Type III

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 $\operatorname{arcsec} x$. This suggests that $\operatorname{arcsec} (2x)$ might just pull it off:

\begin{align*} \left[ \operatorname{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 \, \operatorname{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:

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

### Natural Logarithm

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.

### Polynomial-Logarithms

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 $4x>0$).

### Polynomial-Exponentials

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 $x \in \mathbb{R}$).

### Polynomial-Trigonometrics

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!

### Integration By Parts

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:

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

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

## The Overshooting Method — The Full-Fledged Approach

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.

### $x\sqrt{2x – 3}$

\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*}

### $\dfrac{2x}{3x-1}$

\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*}

### $\dfrac{1}{(2x)^2 + 15}$

\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*}

### $6x \cos(2x^2)$

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*}

### $\ln (5-2x)$

\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*}

### $\arcsin x$

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

### $e^x \cos x$

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*}

### $e^x \sin x$

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*}

### $\sec x$

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!

## Afterword

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 FractionIntegration 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:

1.  $f$ is antidifferentiable, with $F(x)$ being one of its antiderivatives on $I$.
2.  $g$ is differentiable, with $g'(x)$ denoting its derivative on $I$.
3. 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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1. Chirs Polgar says:

Hey. That last one was pure genius. Thanks!

1. 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. 🙂