Integral Calculus Question 278
Question: If $ \int_{{}}^{{}}{\frac{dx}{1+\sin x}=\tan ( \frac{x}{2}+a )+b} $ , then
[Roorkee 1979]
Options:
A) $ a=\frac{\pi }{4},\ b=3 $
B) $ a=-\frac{\pi }{4},\ b=3 $
C) $ a=\frac{\pi }{4},\ b= $ arbitrary constant
D) $ a=-\frac{\pi }{4},\ b= $ arbitrary constant
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{dx}{1+\sin x}}=\tan x-\sec x+c=-\frac{1-\sin x}{\cos x} $ $ =-\frac{{{( \cos \frac{x}{2}-\sin \frac{x}{2} )}^{2}}}{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}+c=-\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}+c $ $ =\frac{\tan \frac{x}{2}-1}{1+\tan \frac{x}{2}}+c=\tan ( \frac{x}{2}-\frac{\pi }{4} )+c $
$ \Rightarrow a=-\frac{\pi }{4},b= $ arbitrary constant.
BETA
