A) \[\frac{1}{4}\log 3\]
B) \[\log 3\]
C) \[\frac{1}{2\log \,3}\]
D) \[2\log 3\]
Correct Answer: A
Solution :
\[\int_{0}^{\pi /4}{\frac{\sin x+\cos x}{3+\sin 2x}}dx\] \[=\int_{0}^{\pi /4}{\frac{(\sin x+\cos x)}{4-(1-\sin 2x)}dx}\] \[=\int_{0}^{\pi /4}{\frac{(\sin x+\cos x)}{4-{{(\sin x-\cos x)}^{2}}}dx}\] Put \[t=(sinx-cosx)\] \[dt=(\cos x+\sin x)dx\] \[=\int_{-1}^{0}{\frac{dt}{(4-{{t}^{2}})}}\] \[=\int_{-1}^{0}{\frac{dt}{(2+t)(2-t)}}\] \[=\frac{1}{4}\int_{-1}^{0}{\left[ \frac{1}{2+t}+\frac{1}{2-t} \right]}dt\] \[=\frac{1}{4}[\log (2+t)-\log (2-t)]_{-1}^{0}\] \[=\frac{1}{4}\left[ \log \left( \frac{2+t}{2-t} \right) \right]_{-1}^{0}\] \[=\frac{1}{4}\left[ \log (1)-log\left( \frac{1}{3} \right) \right]\] \[=\frac{1}{4}\log 3\]
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