A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: C
Solution :
\[\int_{0}^{\pi /2}{\frac{(1+2\cos x)}{{{(2+\cos x)}^{2}}}dx=\int_{0}^{\pi /2}{\frac{2(\cos x+2)-3}{{{(2+\cos x)}^{2}}}dx}}\] \[=2\int_{0}^{\pi /2}{\frac{dx}{2+\cos x}-3\int_{0}^{\pi /2}{\frac{dx}{{{(2+\cos x)}^{2}}}}}\] \[=4\int_{0}^{1}{\frac{dt}{3+{{t}^{2}}}-6\int_{0}^{1}{\frac{1+{{t}^{2}}}{{{(3+{{t}^{2}})}^{2}}}dt}}\], \[\left[ \text{Put}\,\,\tan \frac{x}{2}=t \right]\] \[=-2\int_{0}^{1}{\frac{dt}{3+{{t}^{2}}}+12\int_{0}^{1}{\frac{dt}{{{(3+{{t}^{2}})}^{2}}}}}\] \[=-2\int_{0}^{1}{\frac{dt}{3+{{t}^{2}}}+12}\left[ \frac{1}{6}.\frac{t}{{{t}^{2}}+3} \right]_{0}^{1}+\frac{1}{6}\int_{0}^{1}{\frac{dt}{3+{{t}^{2}}}}\] \[=2\left[ \frac{t}{{{t}^{2}}+3} \right]_{0}^{1}=\frac{1}{2}\].
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