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Q : 19 The area bounded by the $\small y$ -axis, $\small y=\cos x$ and $\small y=\sin x$ when $\small 0\leq x\leq \frac{\pi }{2}$ is

(A) $\small 2(\sqrt{2}-1)$ (B) $\small \sqrt{2}-1$ (C) $\small \sqrt{2}+1$ (D) $\small \sqrt{2}$

Answers (1)

best_answer

Given : $\small y=\cos x$ and $\small y=\sin x$

Area of shaded region = area of BCDB + are of ADCA

$=\int_{\frac{1}{\sqrt{2}}}^{1}x dy +\int_{1}^{\frac{1}{\sqrt{2}}}x dy$

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

$=[y. cos^{-1}y - \sqrt{1-y^2}]_\frac{1}{\sqrt{2}}^1 + [x. sin^{-1}x + \sqrt{1-x^2}]_1^\frac{1}{\sqrt{2}}$

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

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

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

$=\sqrt{2} - 1$

Hence, the correct answer is B.

Posted by

seema garhwal

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