Question:
Evaluate the following integrals:
$\int \frac{\sec ^{2} \sqrt{x}}{\sqrt{x}} d x$
Solution:
Assume $\sqrt{x}=\mathrm{t}$
$d(\sqrt{x})=d t$
$\Rightarrow \frac{1}{2 \sqrt{x}} \mathrm{dx}=\mathrm{dt}$
$\Rightarrow \frac{1}{\sqrt{x}} \mathrm{dx}=2 \mathrm{dt}$
Substituting $t$ and $d t$
$\Rightarrow 2 \int \sec ^{2} t d t$
$=2 \tan t+c$
But $\sqrt{x}=t$
$\Rightarrow 2 \tan (\sqrt{x})+c$
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