A) \[\frac{1}{\sqrt{6}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{6}}\tan \frac{x}{2} \right)+c\]
B) \[\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}}\tan \frac{x}{2} \right)+c\]
C) \[\frac{1}{4}{{\tan }^{-1}}\left( \tan \frac{x}{2} \right)+c\]
D) \[\frac{1}{7}{{\tan }^{-1}}\left( \tan \frac{x}{2} \right)+c\]
Correct Answer: A
Solution :
\[I=\frac{dx}{7+5\cos x}\]\[=\int{\frac{dx}{7+5\,\left( \frac{1-{{\tan }^{2}}(x/2)}{1+{{\tan }^{2}}(x/2)} \right)}}\] \[=\int{\frac{{{\sec }^{2}}(x/2)\,dx}{7+7{{\tan }^{2}}(x/2)+5-5{{\tan }^{2}}(x/2)}}\] \[=\int{\frac{{{\sec }^{2}}(x/2)\,dx}{12+2{{\tan }^{2}}(x/2)}}\]\[=\int{\frac{\frac{1}{2}{{\sec }^{2}}(x/2)\,.dx}{6+{{\tan }^{2}}(x/2)}}\] Put \[\tan \frac{x}{2}=t\] Þ \[\frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt\] \[I=\int{\frac{dt}{{{t}^{2}}+({{\sqrt{6)}}^{2}}}}\]\[=\frac{1}{\sqrt{6}}{{\tan }^{-1}}\frac{t}{\sqrt{6}}+c\] \[=\frac{1}{\sqrt{6}}{{\tan }^{-1}}\left| \frac{\tan (x/2)}{\sqrt{6}} \right|+c\].
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