Evaluate the following integrals:
Question:

Evaluate the following integrals:

$\int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x$

Solution:

let $I=\int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x$

Let $\tan x=t \ldots \ldots(i)$

$\Rightarrow \sec ^{2} x d x=d t$

SO,

$I=\int \frac{d t}{(1)^{2}-t^{2}}$

$\mathrm{I}=\frac{1}{2 \times 1} \log \left|\frac{1+\mathrm{t}}{1-\mathrm{t}}\right|+\mathrm{c}\left[\right.$ since, $\left.\int \frac{1}{\mathrm{a}^{2}-(\mathrm{x})^{2}} \mathrm{dx}=\frac{1}{2 \times \mathrm{a}} \log \left|\frac{\mathrm{a}+\mathrm{x}}{\mathrm{a}-\mathrm{x}}\right|+\mathrm{c}\right]$

$I=\frac{1}{2} \log \left|\frac{1+\tan x}{1-\tan x}\right|+c$ [using (i)]

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