A) \[\frac{\pi }{4}+\frac{1}{2}\log 2\]
B) \[\frac{\pi }{4}+\log 2\]
C) \[\frac{\pi }{4}-\frac{1}{2}\log 2\]
D) \[\frac{\pi }{4}-\log 2\]
Correct Answer: C
Solution :
\[\int_{0}^{\pi /2}{\frac{\cos x}{1+\cos x+\sin x}}dx\] \[=\int_{0}^{\pi /2}{\frac{{{\cos }^{2}}(x/2)-{{\sin }^{2}}(x/2)}{2{{\cos }^{2}}(x/2)+2\sin (x/2)\cos (x/2)}}dx\] \[=\frac{1}{2}\int_{0}^{\pi /2}{\frac{1-{{\tan }^{2}}(x/2)}{1+\tan (x/2)}}dx=\frac{1}{2}\int_{0}^{\pi /2}{\left[ 1-\tan \left( \frac{x}{2} \right) \right]}dx\] \[\frac{\pi }{4}+\log \frac{1}{\sqrt{2}}=\frac{\pi }{4}-\frac{1}{2}\log 2\].
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