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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Prove $\int_{0}^{1} \sin ^{-1} x\  d\  x=\frac{\pi}{2}-1$
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Answer

Given integral is: $\int_{0}^{1} \sin ^{-1} x\ d\ x$
To Prove: $\int_{0}^{1} \sin ^{-1} x\ d\ x=\frac{\pi}{2}-1$
Let $I=\int_{0}^{1} \sin ^{-1} x \cdot 1 d\ x$
because, $\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} \cdot \int \mathrm{v} \mathrm{dx}-\int \frac{\mathrm{d}}{\mathrm{dx}} u \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow I$ = $\sin ^{-1} \mathrm{x} \cdot \int_{0}^{1} 1 \cdot \mathrm{dx}-\int_{0}^{1} \frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x} \cdot\left\{\int 1 \cdot \mathrm{dx}\right\} \cdot \mathrm{dx}$
$\Rightarrow I=\left[\sin ^{-1} \mathrm{x} \cdot \mathrm{x}\right]_{0}^{1}-\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \cdot \mathrm{x} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\left[\sin ^{-1} \mathrm{x} \cdot \mathrm{x}\right]_{0}^{1}-\mathrm{I}_1$ .....(i)
First solving $I_1:$
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \cdot \mathrm{x} \mathrm{d} \mathrm{x}$
Let $1 - x^2 = t $
$\Rightarrow -2 x dx = dt$
When $x = 0$ then $t = 1$ and when $x = 1$ then $t = 0$
$\Rightarrow \mathrm{I}_{1}=\int_{1}^{0} \frac{1}{\sqrt{\mathrm{t}}} \cdot \frac{-\mathrm{dt}}{2}$
$= -\frac{1}{2}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right]_{1}^{0}$
$\Rightarrow \mathrm{I}_{1}=\sqrt{1}$
$\Rightarrow \mathrm{I}_{1}=1$
Put in equation $(i)$
$\Rightarrow \mathrm{I}=\left[\sin ^{-1} \mathrm{x} \cdot \mathrm{x}\right]_{0}^{1}-1$
$\Rightarrow \mathrm{I}=\sin ^{-1}(1)-0-1$
$\Rightarrow \mathrm{I}=\frac{\pi}{2}-1$
Hence Proved.

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