Integral Calculus Question 7
Question: If $ \int{\frac{\cos 4x+1}{\cos x-\tan x}}dx=k\cos 4x+c $ then
[DCE 2005]
Options:
A) $ k=-1/2 $
B) $ k=-1/8 $
C) $ k=-1/4 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int{\frac{1+\cos 4x}{\cot x-\tan x}}dx=\int{\frac{2{{\cos }^{2}}2x}{{{\cos }^{2}}x-{{\sin }^{2}}x}}.\sin x\cos xdx $ $ =\int{\cos 2x\sin 2xdx=-\frac{1}{8}\cos 4x+c} $ ,
$ \therefore $ $ k=-1/8 $ .
BETA
