Integral Calculus Question 412
Question: $ \int_{{}}^{{}}{\cos 2\theta \log ( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } )\ d\theta =} $
[IIT 1994]
Options:
A) $ {{(\cos \theta -\sin \theta )}^{2}}\log ( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } ) $
B) $ {{(\cos \theta +\sin \theta )}^{2}}\log ( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } ) $
C) $ \frac{{{(\cos \theta -\sin \theta )}^{2}}}{2}\log ( \frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta } ) $
D) $ \frac{1}{2}\sin 2\theta \log \tan ( \frac{\pi }{4}+\theta )-\frac{1}{2}\log \sec 2\theta $
Show Answer
Answer:
Correct Answer: D
Solution:
We know that
$ \log ( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } )=\log ( \frac{1+\tan \theta }{1-\tan \theta } )=\log \tan ( \frac{\pi }{4}+\theta ) $
$ \int_{{}}^{{}}{\sec \theta ,d\theta }=\log \tan ( \frac{\pi }{4}+\frac{\theta }{2} ) $
$ \therefore ,\int_{{}}^{{}}{\sec 2\theta ,d\theta }=\frac{1}{2}\log \tan ( \frac{\pi }{4}+\theta ) $
$ \therefore ,2\sec 2\theta =\frac{d}{d\theta }\log \tan ( \frac{\pi }{4}+\theta ) $ .?.(i)
Integrating the given expression by parts, we get
$ I=\frac{1}{2}\sin 2\theta \log \tan ( \frac{\pi }{4}+\theta )-\frac{1}{2}\int_{{}}^{{}}{\sin 2\theta ,.,2\sec 2\theta ,d\theta } $ by (i)
$ =\frac{1}{2}\sin 2\theta \log \tan ( \frac{\pi }{4}+\theta )-\int_{{}}^{{}}{\tan 2\theta },d\theta $
$ =\frac{1}{2}\sin 2\theta \log \tan ( \frac{\pi }{4}+\theta )-\frac{1}{2}\log \sec 2\theta $ .
BETA
