MCQ
$0 < \alpha < \pi $ માટે $\int_0^1 {\frac{{dx}}{{{x^2} + 2x\cos \alpha + 1}}} =$
- A$\sin \alpha $
- B${\tan ^{ - 1}}(\sin \alpha )$
- C$\alpha \sin \alpha $
- ✓$\frac{\alpha }{2}{(\sin \alpha )^{ - 1}}$
$ = \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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