A) \[\pi /4\]
B) \[\pi /8\]
C) \[\pi /2\]
D) zero
Correct Answer: A
Solution :
\[\int_{0}^{\pi }{\frac{dx}{5+3\cos x}}=\int_{0}^{\pi }{\frac{dx}{5+3\left\{ \frac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2} \right\}}}\] \[=\int_{0}^{\pi }{\frac{{{\sec }^{2}}x/2dx}{5+5{{\tan }^{2}}x/2+3-3{{\tan }^{2}}x/2}}\] \[=\int_{0}^{\pi }{\frac{{{\sec }^{2}}x/2dx}{8+2{{\tan }^{2}}x/2}}\] \[=\frac{1}{2}\int_{0}^{\pi }{\frac{{{\sec }^{2}}x/2dx}{{{(2)}^{2}}{{(\tan x/2)}^{2}}}}\tan x/2=t\] \[{{\sec }^{2}}x/2dx=2dt\] \[=\int_{0}^{\infty }{\frac{dt}{{{(2)}^{2}}+{{(t)}^{2}}}}=\left[ \frac{1}{2}{{\tan }^{-1}}\frac{t}{2} \right]_{0}^{\infty }\] \[=\frac{1}{2}\left[ \frac{\pi }{2}-0 \right]=\pi /4\]
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