A) \[\frac{1}{20}\log 3\]
B) \[\log 3\]
C) \[\frac{1}{20}\log 5\]
D) None of these
Correct Answer: A
Solution :
Let \[I=\int_{0}^{\pi /4}{\frac{\sin x+\cos x}{9+16\sin 2x}\,}dx\] Put \[\sin x-\cos x=t\], then \[(\sin x+\cos x)dx=dt\] \[I=\int_{-1}^{0}{\frac{dt}{9+16(1-{{t}^{2}})}}=\int_{-1}^{0}{\frac{dt}{25-16{{t}^{2}}}}\] \[=\frac{1}{10}\int_{-1}^{0}{\left( \frac{1}{5-4t}+\frac{1}{5+4t} \right)dt}\] \[=\left| \frac{1}{10}.\frac{1}{4}[\log (5+4t)-\log (5-4t)]\, \right|_{-1}^{0}\] \[=\frac{1}{40}(\log 9-\log 1)=\frac{1}{20}\log 3\].
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