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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.11
By using the properties of definite integrals, evaluate the integrals in Exercises 208 to 225.
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Question : 216 of 271
Marks: +1, -0
∫0π/2(2log⁡sin⁡x−log⁡sin⁡2x)dx\int\limits_{0}^{\pi/2}(2\log \sin x-\log \sin 2x)dx
Solution:  
Let I=∫0π/2(2log⁡sin⁡x−log⁡sin⁡2x)dxI=\int\limits_{0}^{\pi/2}(2\log \sin x-\log \sin 2x)dx
=∫0π/2[2log⁡sin⁡x−log⁡(2sin⁡xcos⁡x)]dx=\int\limits_{0}^{\pi/2}[2\log \sin x-\log (2\sin x\cos x)]dx
=∫0π/2[log⁡sin⁡x−log⁡2−log⁡cos⁡x]dx=\int\limits_{0}^{\pi/2}[ \log \sin x-\log 2-\log\cos x]dx
=∫0π/2log⁡sin⁡xdx−∫0π/2log⁡2dx−∫0π/2log⁡cos⁡xdx=\int\limits_{0}^{\pi/2} \log \sin xdx-\int\limits_{0}^{\pi/2}\log 2dx -\int\limits_{0}^{\pi/2}\log\cos x dx
=∫0π/2log⁡sin⁡xdx−∫0π/2log⁡2dx−∫0π/2log⁡cos⁡(π/2−x)dx=\int\limits_{0}^{\pi/2} \log \sin xdx-\int\limits_{0}^{\pi/2}\log 2dx -\int\limits_{0}^{\pi/2}\log\cos(\pi/2- x) dx
[∵∫0af(x)dx=∫0af(a−x)dx]\left[\because \int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx\right]
=∫0π/2log⁡sin⁡xdx−(log⁡2)[x]0π/2−∫0π/2log⁡sin⁡xdx=\int\limits_{0}^{\pi/2} \log \sin xdx- (\log 2)[x]_{0}^{\pi/2} -\int\limits_{0}^{\pi/2}\log\sin x dx
=−(log⁡2)(π/2−0)=-(\log 2)(\pi/2-0)
=−π/2log⁡2=-\pi/2 \log 2
=π/2log⁡(1/2)=\pi/2 \log (1/2)
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