Integral Calculus Question 202
Question: If $ \int{\frac{1}{1+\sin x}dx=\tan ( \frac{x}{2}+a )+b} $ then
Options:
A) $ a=-\frac{\pi }{4},b\in R $
B) $ a=\frac{\pi }{4},b\in R $
C) $ a=\frac{5\pi }{4},b\in R $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ I=\int{\frac{1}{1+\sin x}dx=\int{\frac{dx}{1+\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}}} $ $ \int{\frac{( 1+{{\tan }^{2}}\frac{x}{2} )dx}{1+{{\tan }^{2}}\frac{x}{2}+2\tan \frac{x}{2}}=\int{\frac{{{\sec }^{2}}\frac{x}{2}dx}{1+{{\tan }^{2}}\frac{x}{2}+2\tan \frac{x}{2}}}} $ Substitute $ \tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt $
$ \Rightarrow {{\sec }^{2}}\frac{x}{2}dx=2dt $ . Then $ I=\int{\frac{2dt}{1+t^{2}+2t}=2\int{\frac{dt}{{{(1+t)}^{2}}}=2\frac{-1}{(1+t)}+C}} $ $ =\frac{-2}{1+\tan \frac{x}{2}}+c $ $ =1-\frac{2}{1+\tan \frac{x}{2}}+(c-1)=\frac{\tan \frac{x}{2}-1}{\tan \frac{x}{2}+1}+b $ , Where $ b=c-1 $ , a new constant $ =-\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}+b=-\tan ( \frac{\pi }{4}-\frac{\pi }{2} )+b $ $ =\tan ( \frac{\pi }{2}-\frac{\pi }{4} )+b $ Clearly $ a=-\frac{\pi }{4} $ and $ b\in R $
BETA
