Integral Calculus Question 316
Question: $ \int_{{}}^{{}}{x^{2}\sin 2x}\ dx= $
[IIT 1974]
Options:
A) $ \frac{1}{2}x^{2}\cos 2x+\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x+c $
B) $ -\frac{1}{2}x^{2}\cos 2x+\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x+c $
C) $ \frac{1}{2}x^{2}\cos 2x-\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ I=\int_{{}}^{{}}{x^{2}\sin 2x,dx}=\frac{-x^{2}\cos 2x}{2}+\int_{{}}^{{}}{\frac{2x\cos 2x}{2},dx}+c $ $ =-\frac{x^{2}\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+c. $
BETA
