Show that

$\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1$


Let $I=\int_{0}^{1} \sin ^{-1} x d x$

$\Rightarrow I=\int_{0}^{1} \sin ^{-1} x \cdot 1 \cdot d x$

Integrating by parts, we obtain

$I=\left[\sin ^{-1} x \cdot x\right]_{0}^{1}-\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} \cdot x d x$

$=\left[x \sin ^{-1} x\right]_{0}^{1}+\frac{1}{2} \int_{0}^{1} \frac{(-2 x)}{\sqrt{1-x^{2}}} d x$

Let $1-x^{2}=t \Rightarrow-2 x d x=d t$

When $x=0, t=1$ and when $x=1, t=0$

$I=\left[x \sin ^{-1} x\right]_{0}^{1}+\frac{1}{2} \int_{1}^{0} \frac{d t}{\sqrt{t}}$

$=\left[x \sin ^{-1} x\right]_{0}^{1}+\frac{1}{2}[2 \sqrt{t}]_{1}^{0}$

$=\sin ^{-1}(1)+[-\sqrt{1}]$


Hence, the given result is proved.





Leave a comment

Please enter comment.
Please enter your name.