Question

The area bounded by $y={\sin }^{-1}x$, $y=\frac{\pi }{4}$ and $Y$-axis is

• A. $\frac{1}{\sqrt{2}}+1$
• B. $1-\frac{1}{\sqrt{2}}$
• C. $\frac{\pi }{4\sqrt{2}}$
• D. $\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

Answer 1

The correct option is D $\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

Required area $=$ Area of rectangle $OABC-$ Area of curve $OABO$
$=\frac{\pi }{4}\times \frac{1}{\sqrt{2}}-{\int }_0^{\pi /4}\sin ydy$
$=\frac{\pi }{4\sqrt{2}}+{\left[\cos y\right]}_0^{\pi /4}$
$=\frac{\pi }{4\sqrt{2}}+\cos \frac{\pi }{4}-\cos 0$
$=\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1$