A) \[\sin \alpha \]
B) \[{{\tan }^{-1}}(\sin \alpha )\]
C) \[\alpha \sin \alpha \]
D) \[\frac{\alpha }{2}{{(\sin \alpha )}^{-1}}\]
Correct Answer: D
Solution :
\[\int_{0}^{1}{\frac{dx}{{{x}^{2}}+2x\cos \alpha +1}=\int_{0}^{1}{\frac{dx}{{{(x+\cos \alpha )}^{2}}+1-{{\cos }^{2}}\alpha }}}\] \[=\int_{0}^{1}{\frac{dx}{{{(x+\cos \alpha )}^{2}}+{{\sin }^{2}}\alpha }=\left[ \frac{1}{\sin \alpha }{{\tan }^{-1}}\frac{x+\cos \alpha }{\sin \alpha } \right]}_{0}^{1}\] \[=\frac{1}{\sin \alpha }\left( {{\tan }^{-1}}\cot \frac{\alpha }{2}-{{\tan }^{-1}}\cot \alpha \right)=\frac{\alpha }{2}.\frac{1}{\sin \alpha }\].
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