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Question: Answered & Verified by Expert
$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} d x=$
MathematicsDefinite IntegrationMHT CETMHT CET 2022 (08 Aug Shift 2)
Options:
  • A $\sqrt{2}+1$
  • B $1+2 \sqrt{2}$
  • C $\sqrt{2}-1$
  • D $2 \sqrt{2}-1$
Solution:
2302 Upvotes Verified Answer
The correct answer is: $\sqrt{2}-1$
$\begin{aligned} & \int_0^{\pi / 4} \sqrt{1-\sin 2 x} \mathrm{~d} x=\int_0^{\pi / 4} \sqrt{\cos ^2 x+\sin ^2 x-2 \sin x \cdot \cos x} \mathrm{~d} x \\ & =\int_0^{\pi / 4} \sqrt{(\cos x-\sin x)^2} \mathrm{~d} x=\int_0^{\pi / 4}|\cos x-\sin x| \mathrm{d} x \\ & =\int_0^{\pi / 4}(\cos x-\sin x) \mathrm{d} x=[\sin x+\cos x]_0^{\pi / 4} \\ & =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1)=-1\end{aligned}$

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