Evaluate the following integrals:
Question:

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

Solution:

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

$\int u \cdot d v=u v-\int v d u$

Choose $u$ in these order LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG, E-EXPONENTIAL)

Here $u=\sec ^{-1} \sqrt{x}$ and $v=1$

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

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

Put $x-1=t d x=d t$

$=x \sec ^{-1} x-\int \frac{d t}{2 \sqrt{t}}$

$=x \sec ^{-1} x-\frac{2}{2}(\sqrt{t})+c$

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

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