_ 4 ^ 4 _e ( x + x ) ,dx is equal to - Vidyadip

MCQ

$\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{4}} {{{\log }_e}\left( {\sin x + \cos x} \right)\,dx}$ is equal to

✓
$ - \frac{\pi }{4}\ln 2$
B
$ \frac{\pi }{4}\ln 2$
C
$ \frac{\pi }{8}\ln 2$
D
$ -\frac{\pi }{8}\ln 2$

✓

Answer

Correct option: A.

$ - \frac{\pi }{4}\ln 2$

a
$I=\int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \log \left\{\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\right\} d x$

Putting $\left(\mathrm{x}+\frac{\pi}{4}\right)=\theta ; \mathrm{dx}=\mathrm{d} \theta$

$=\int_{0}^{\frac{\pi}{2}} \log (\sqrt{2} \sin \theta) d \theta$

$=\int_{0}^{\frac{\pi}{2}} \log \sqrt{2} \mathrm{d} \theta+\int_{0}^{\frac{\pi}{2}} \log \sin \theta \mathrm{d} \theta$

$=\log \sqrt{2}(\theta)_{0}^{\pi / 2}-\frac{\pi}{2} \ln 2$

$=\frac{\pi}{4} \ln 2-\frac{\pi}{2} \ln 2=-\frac{\pi}{4} \ln 2$

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