A) \[\frac{\pi }{2(1-{{a}^{2}})}\]
B) \[\pi (1-{{a}^{2}})\]
C) \[\frac{\pi }{1-{{a}^{2}}}\]
D) None of these
Correct Answer: C
Solution :
\[\int_{0}^{\pi }{\frac{dx}{(1+{{a}^{2}})\left( {{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2} \right)-2a\left( {{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2} \right)}}\] \[=\int_{0}^{\pi }{\frac{dx}{{{(1-a)}^{2}}{{\cos }^{2}}\frac{x}{2}+{{(1+a)}^{2}}{{\sin }^{2}}\frac{x}{2}}}\] \[=\frac{2}{{{(1+a)}^{2}}}\int_{0}^{\infty }{\frac{dt}{{{\left\{ (1-a)/(1+a) \right\}}^{2}}+{{t}^{2}}}}\]; {where \[t=\tan \frac{x}{2}\]} \[=\frac{2}{{{(1+a)}^{2}}}\frac{(1+a)}{(1-a)}\left[ {{\tan }^{-1}}\left( \frac{1+a}{1-a}.t \right) \right]_{0}^{\infty }\] \[=\frac{2}{(1-{{a}^{2}})}[{{\tan }^{-1}}\infty -{{\tan }^{-1}}0]=\frac{\pi }{1-{{a}^{2}}}\].
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