Science:Math Exam Resources/Courses/MATH101 A/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:

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

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,

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