Integrating Derivatives of Inverse Trig Functions

(Videos) Integration Using Inverse Trigonometric Functions

by PatrickJMT

Introduction

In this video, Patrick shows you how to solve integration problems using inverse trigonometric functions. We all remember the derivatives of inverse trigonometric functions right? No? Well here's a recap.

Well one integration technique is noticing these derivatives inside an integral and reverting them back into their inverse trig form.

The two most popular ones that Patrick looks at are: $$\int\frac{1}{\sqrt{1-x^2}}\,dx=\sin^{-1}x+C\\ \int\frac{1}{1+x^2}\,dx=\tan^{-1}x+C$$

**Example **

The example that Patrick looks at is: $$\int\frac{1}{\sqrt{1-4t^2}}\,dt$$

The first step is to figure out what inverse trig derivative it looks most like. Here. our problem looks similar to: $$\color{red}{\int\frac{1}{\sqrt{1-x^2}}\,dx}$$ which is equal to $\sin^{-1}(x)$.

The next step is to get our integral to look like the common one. This can usually be done with a little bit of algebra or $u$-substitution. In this example we have a 4 in front of the $t^2$ which isn't letting us integrate easily. $$\int\frac{1}{\sqrt{1-\color{red}{4}t^2}}\,dt$$ But what if we rewrote the expression $4t^2$ as $(2t)^2$. Now we have some expression being squared and no values out front.

Let's use $u$-substitution now. Let $u=2t$. Then, $du=2\,dt$ or $\frac12\,du=dt$.

$$\begin{align} \int\frac{1}{\sqrt{1-4t^2}}\,dt&=\int\frac{1}{\sqrt{1-(2t)^2}}\,dt\\ &=\int\frac{1}{\sqrt{1-u^2}}\left(\frac12\right)\,du\\ &=\frac12\int\frac{1}{\sqrt{1-u^2}}\,du\\ &=\frac12\sin^{-1}(u)+C\\ &=\boxed{\frac12\sin^{-1}(2t)+C} \end{align}$$

Example

In this video, Patrick looks at another example. This one being: $$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\cos^2x}\,dx$$

Let's rewrite this integral in order to see how it looks like an inverse trig derivative. $$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\left(\cos x\right)^2}\,dx$$

This looks similar to the inverse tangent derivative: $$\color{red}{\int\frac{1}{1+x^2}\,dx}$$

Let's start with some $u$-substitution. Let $u=\cos x$. Then, $du=-\sin x\,dx$ or $-du=\sin x\,dx$. Our integral becomes: $$\begin{align} \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\left(\cos x\right)^2}\,dx&=-\int_0^{\frac{\pi}{2}}\frac{1}{1+\left(u\right)^2}\,du\\ &=-\left[\tan^{-1}(u)\right]_0^{\frac{\pi}{2}}\\ &=-\left[\tan^{-1}(\cos x)\right]_0^{\frac{\pi}{2}}\\ &=-\left[\tan^{-1}\left(\cos \frac{\pi}2\right)-\tan^{-1}\left(\cos 0\right)\right]\\ &=-\left[\tan^{-1}(0)-\tan^{-1}(1)\right]\\ &=-\left[0-\frac{\pi}4\right]\\ &=\frac{\pi}4\\ \end{align}$$

It is important to keep in mind the limits of integration when using $u$-substitution. You cannot plug the original limits of integration into the $u$ form. You must either convert the $u$ form back to its original form and use the limits, or change the limits to their $u$ form. In this problem I kept the limits of integration the same because I like to revert everything back and then use them.