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Home  >>  CBSE XII  >>  Math  >>  Integrals
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$\Large \int\limits_0^\frac{\pi}{2}\normalsize \sqrt {1-\sin2x}\;dx$ is equal to $(A)\;0\quad(B)\;2\quad(C)\;1\quad(D)\;-1$

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  • $\sin^2x+\cos^2x=1$
  • $\int\sin xdx=-\cos x+c.$
  • $\int \cos xdx=\sin x+c.$
Step 1:
Let $I=\int_0^{\Large\frac{\pi}{2}}\sqrt{1-\sin2x}dx.$
We know $\sin^2x+\cos^2x=1$ and $\sin 2x=2\sin x\cos x$
$1-\sin 2x$ can be written as $(\sin^2x+\cos^2x-2\sin x\cos x).$
This is of the form $(a-b)^2$
(i.e)$(\sin x-\cos x)^2$
Hence $\sqrt{1-\sin 2x}=\sqrt{(\sin x-\cos x)^2}=(\sin x-\cos x)$
Step 2:
$I=\int_0^{\Large\frac{\pi}{2}}(\sin x-\cos x)dx.$
On integrating we get,
$[-\cos x-\sin x]_0^{\large\frac{\pi}{2}}=-[\cos x+\sin x]_0^{\large\frac{\pi}{2}}$
Step 3:
On applying limits we get,
$I=-[(\cos\large\frac{\pi}{2}$$-\cos 0)+(\sin\large\frac{\pi}{2}$$-\sin 0)]$
$\cos\large\frac{\pi}{2}$$=\sin 0=0$ and $\sin\large\frac{\pi}{2}$$=\cos 0=1.$
Hence the correct option is (A)
answered Apr 25, 2013 by sreemathi.v

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