# The area bounded by the

Question:

The area bounded by the $y$-axis, $y=\cos x$ and $y=\sin x$ when $0 \leq x \leq \frac{\pi}{2}$

A. $2(\sqrt{2}-1)$

B. $\sqrt{2}-1$

C. $\sqrt{2}+1$

D. $\sqrt{2}$

Solution:

The given equations are

y = cos x … (1)

And, y = sin x … (2)

Required area = Area (ABLA) + area (OBLO)

$=\int_{\frac{1}{\sqrt{2}}}^{1} x d y+\int_{0}^{\frac{1}{\sqrt{2}}} x d y$

$=\int_{\frac{1}{\sqrt{2}}}^{1} \cos ^{-1} y d y+\int_{0}^{\frac{1}{\sqrt{2}}} \sin ^{-1} x d y$

Integrating by parts, we obtain

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

$=\left[\cos ^{-1}(1)-\frac{1}{\sqrt{2}} \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-\frac{1}{2}}\right]+\left[\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-\frac{1}{2}}-1\right]$

$=\frac{-\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1$

$=\frac{2}{\sqrt{2}}-1$

$=\sqrt{2}-1$ units

Thus, the correct answer is B.