Answer
$$\int_{0}^{\frac{\pi}{4}} \tan x \sec ^{2} x d x =\frac{1}{2}$$
Work Step by Step
Given :
\[
\int_{0}^{\frac{\pi}{4}} \sec ^{2} x \tan x d x
\]
Let us consider
$\Longrightarrow d u=\tan x \sec x d x \quad$
For $u=1, x=0$, $x=\frac{\pi}{4}, \sqrt{2}=u$ Putting these values in the given integral, we have
\[
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \tan x \sec ^{2} x d x &=\int_{0}^{\frac{\pi}{4}} [\tan x \sec x d x]\sec x \\
&=\int_{1}^{\sqrt{2}} u d u \\
&=\left[\frac{u^{2}}{2}\right]_{1}^{\sqrt{2}} \\
&=\left[-\frac{1}{2}+1\right] \\
&=\frac{1}{2}
\end{aligned}
\]
Let us consider
\[
\tan=u
\]
$\Longrightarrow d u=\sec ^{2} x d x \quad$
For $u=0, x=0$, $x=\frac{\pi}{4}, u=1$
\[
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \tan x \sec ^{2} x d x &=\int_{0}^{1} u d u \\
&=\left[\frac{u^{2}}{2}\right]_{0}^{1} \\
&=\left[-0+\frac{1}{2} \right] \\
&=\frac{1}{2}
\end{aligned}
\]
