Integral Calculus Question 348
Question: $ \int_{{}}^{{}}{2\sin x}\cos x\ dx $ is equal to
[SCRA 1996]
Options:
A) $ \cos 2x+c $
B) $ \sin 2x+c $
C) $ {{\cos }^{2}}x+c $
D) $ {{\sin }^{2}}x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_{{}}^{{}}{2\sin x,.,\cos x,dx}=\int_{{}}^{{}}{\sin 2x,dx} $ $ =-\frac{\cos 2x}{2}+c=-\frac{(1-2{{\sin }^{2}}x)}{2}+c $ $ =-\frac{1}{2}+{{\sin }^{2}}x+c={{\sin }^{2}}x+c $ .
BETA
