Integral Calculus Question 139
Question: $ \int_{{}}^{{}}{x{{\cos }^{2}}}xdx= $
[IIT 1972]
Options:
A) $ \frac{x^{4}}{4}-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c $
B) $ \frac{x^{2}}{4}+\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+c $
C) $ \frac{x^{4}}{4}-\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+c $
D) $ \frac{x^{4}}{4}+\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{x{{\cos }^{2}}x,dx}=\frac{1}{2}\int_{{}}^{{}}{x(1+\cos 2x),dx} $ $ =\frac{x^{2}}{4}+\frac{1}{2}[ \frac{x\sin 2x}{2}-\int_{{}}^{{}}{\frac{\sin 2x}{2},dx} ]+c $ $ =\frac{x^{2}}{4}+\frac{x\sin 2x}{4}+\frac{\cos 2x}{8}+c. $
BETA
