Science:Math Exam Resources/Courses/MATH101 C/April 2025/Question 02 (c)/Solution 1

We simplify the integral by the following trigonometric substitution; let $x=\tan \theta$, so $dx={\sec }^2\theta d\theta .$ Using this substitution, the integral can be rewritten in terms of $\theta$ as follows:

$$\begin{array}{rl}\int \frac{1}{{\tan }^2\theta \sqrt{1+{\tan }^2\theta }}{\sec }^2\theta d\theta & =\int \frac{1}{{\tan }^2\theta \sqrt{{\sec }^2\theta }}{\sec }^2\theta d\theta \\ & =\int \frac{1}{{\tan }^2\theta \sec \theta }{\sec }^2\theta d\theta \\ & =\int \frac{1}{{\tan }^2\theta }\sec \theta d\theta \\ & =\int \frac{{\cos }^2\theta }{{\sin }^2\theta }\frac{1}{\cos \theta }d\theta \\ & =\int \frac{\cos \theta }{{\sin }^2\theta }d\theta .\\ \end{array}$$

Now we do the standard $u-$substitution that takes advantage of the relationship between $\sin$ and $\cos$: $u=\sin \theta ,du=\cos \theta d\theta$. We can then integrate and reverse substitution as follows,

$$\begin{array}{rl}\int \frac{\cos \theta }{{\sin }^2\theta }d\theta & =\int \frac{1}{u^2}du\\ & =\int u^{-2}du\\ & =-\frac{1}{u}+C\\ & =-\frac{1}{\sin \theta }+C\\ & =-\frac{\sqrt{1+x^2}}{x}+C.\end{array}$$