Integral Calculus Question 262
Question: $ \int_{{}}^{{}}{x{{\sin }^{2}}x\ dx=} $
[BIT Ranchi 1977; IIT 1972]
Options:
A) $ \frac{x^{2}}{4}+\frac{x}{4}\sin 2x+\frac{1}{8}\cos 2x+c $
B) $ \frac{x^{2}}{4}-\frac{x}{4}\sin 2x+\frac{1}{8}\cos 2x+c $
C) $ \frac{x^{2}}{4}+\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c $
D) $ \frac{x^{2}}{4}-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{x{{\sin }^{2}}x,dx}=\int_{{}}^{{}}{x,.,\frac{(1-\cos 2x)}{2},dx} $ $ =\frac{1}{2}[ \int_{{}}^{{}}{x,dx}-\int_{{}}^{{}}{x,.,\cos 2x,dx} ]=\frac{x^{2}}{4}-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c $ .
BETA
