Definite Integration Question 386
Question: The value of the definite integral $ \int_0^{1}{\frac{dx}{x^{2}+2x\cos \alpha +1}} $ for $ 0<\alpha <\pi $ is equal to
[Kurukshetra CEE 2002]
Options:
A) $ \sin \alpha $
B) $ {{\tan }^{-1}}(\sin \alpha ) $
C) $ \alpha \sin \alpha $
D) $ \frac{\alpha }{2}{{(\sin \alpha )}^{-1}} $
Show Answer
Answer:
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 }=[ \frac{1}{\sin \alpha }{{\tan }^{-1}}\frac{x+\cos \alpha }{\sin \alpha } ]}_0^{1} $
$ =\frac{1}{\sin \alpha }( {{\tan }^{-1}}\cot \frac{\alpha }{2}-{{\tan }^{-1}}\cot \alpha )=\frac{\alpha }{2}.\frac{1}{\sin \alpha } $ .
BETA
