Integral Calculus Question 216
Question: What is $ \int{\sin x\log (\tan x)dx} $ equal to?
Options:
A) $ \cos x\log \tan x+\log \tan (x/2)+c $
B) $ -\cos x\log \tan x+\log \tan (x/2)+c $
C) $ \cos x\log \tan x+\log \cot ,(x/2)+c $
D) $ -\cos x\log \tan x+\log \cot ,(x/2)+c $
Show Answer
Answer:
Correct Answer: A
Solution:
[b] $ \int{\sin x\log (\tan x)dx} $ $ =-\cos x\log \tan x-\int{(-cosx)\frac{1}{\tan x}.{{\sec }^{2}}xdx} $ $ =-\cos x\log \tan x+\int{\frac{1}{\sin x}dx} $ $ =-\cos x\log (\tan x)+\int{\frac{1+{{\tan }^{2}}\frac{x}{2}}{2\tan \frac{x}{2}}dx} $ Now putting $ \frac{x}{2}=t, $ we get, $ =-\cos x\log \tan x+\int{\frac{1}{t}.dt} $ $ =-\cos x\log \tan x+\log (t)+c $ $ =-\cos x\log \tan x+\log tan( \frac{x}{2} )+c $
BETA
