Integral Calculus Question 85
Question: $ \int \frac{\cos 4x-1}{\cot x-\tan x}dx $ is equal to
Options:
A) $ \frac{1}{2}\ln | \sec 2x |-\frac{1}{4}{{\cos }^{2}}2x+c $
B) $ \frac{1}{2}\ln | \sec 2x |+\frac{1}{4}{{\cos }^{2}}x+c $
C) $ \frac{1}{2}\ln | \cos 2x |-\frac{1}{4}{{\cos }^{2}}2x+c $
D) $ \frac{1}{2}\ln | \cos 2x |+\frac{1}{4}{{\cos }^{2}}x+c $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I=\int{\frac{\cos 4x-1}{\cot x-\tan x}dx} $ $ =\int{\frac{-2{{\sin }^{2}}2x(\sin x,\cos x)}{({{\cos }^{2}}x-{{\sin }^{2}}x)}dx} $ $ =-\int{\frac{{{\sin }^{2}}2x\sin 2x}{\cos 2x}x} $ $ =\int{\frac{({{\cos }^{2}}2x-1)\sin 2x}{\cos 2x}dx} $ Let $ t=\cos 2x $ or $ dt=-2\sin 2xdt $
$ \therefore I=\frac{1}{2}\int{\frac{(1-t^{2})}{t}}dt=\frac{1}{2}\ln | t |-\frac{t^{2}}{4}+C $ $ =\frac{1}{2}\ln | \cos 2x |-\frac{1}{4}{{\cos }^{2}}2x+c $
BETA
