Integral Calculus Question 211
Question: $ \int{\sin 2x.\log \cos xdx} $ is equal to:
Options:
A) $ {{\cos }^{2}}x( \frac{1}{2}+\log \cos x )+k $
B) $ {{\cos }^{2}}x.\log \cos x+k $
C) $ {{\cos }^{2}}x( \frac{1}{2}-\log \cos x )+k $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I=\int{2\sin x.\cos x.\log \cos xdx} $ Put $ \log \cos x=t $
$ \therefore -\frac{\sin x}{\cos x}dx=dt $ $ I=\int{2\sin x.\cos x.t\frac{\cos x}{-\sin x}dt} $ $ =-2\int{{{\cos }^{2}}x.tdt=-2\int{te^{2t}dt}} $ $ =-2[ t.\frac{e^{2t}}{2}-\int{\frac{e^{2t}}{2}.dt} ]=-te^{2t}+\frac{1}{2}e^{2t}+k $ $ e^{2t}( \frac{1}{2}-t )+k={{\cos }^{2}}x.{ \frac{1}{2}-\log \cos x }+k $
BETA
