# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \sec ^{-1} \sqrt{x} d x$

Solution:

Let $\mathrm{I}=\int \sec ^{-1} \sqrt{\mathrm{x}} \mathrm{dx}$

$\sqrt{x}=t ; x=t^{2}$

$\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$

$I=\int 2 t s e c^{-1} t d t$

Using integration by parts,

$=2\left[\sec ^{-1} \mathrm{t} \int \mathrm{tdt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{sec}^{-1} \mathrm{t} \int \mathrm{tdt}\right]$

We know that, $\frac{\mathrm{d}}{\mathrm{dt}} \sec ^{-1} \mathrm{t}=\frac{1}{\mathrm{t} \sqrt{\mathrm{t}^{2}-1}}$

$=2\left[\frac{\mathrm{t}^{2}}{2} \sec ^{-1} \mathrm{t}-\int \frac{1}{\mathrm{t} \sqrt{\mathrm{t}^{2}-1}} \int \mathrm{tdt}\right]$

$=2\left[\frac{\mathrm{t}^{2}}{2} \mathrm{sec}^{-1} \mathrm{t}-\int \frac{\mathrm{t}^{2}}{2 \mathrm{t} \sqrt{\mathrm{t}^{2}-1}} \mathrm{dt}\right]$

$=\mathrm{t}^{2} \sec ^{-1} \mathrm{t}-\int \frac{\mathrm{t}}{\mathrm{t} \sqrt{\mathrm{t}^{2}-1}} \mathrm{dt}$

$=\mathrm{t}^{2} \sec ^{-1} \mathrm{t}-\frac{1}{2} \int \frac{2 \mathrm{t}}{\sqrt{\mathrm{t}^{2}-1}} \mathrm{dt}$

$=\mathrm{t}^{2} \sec ^{-1} \mathrm{t}-\frac{1}{2} \times 2 \sqrt{\mathrm{t}^{2}-1}+\mathrm{c}$

Substitute value for $t$,

$I=x \sec ^{-1} \sqrt{x}-\sqrt{x-1}+c$

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