Integral Calculus Question 37
Question: $ \int_{{}}^{{}}{\sin 2x\cos 3x\ dx=} $
[Roorkee 1976]
Options:
A) $ \frac{1}{2}( \cos x+\frac{1}{5}\cos 5x )+c $
B) $ \frac{1}{2}( \cos x-\frac{1}{5}\cos 5x )+c $
C) $ \cos x+\frac{1}{5}\cos 5x+c $
D) $ \cos x-\frac{1}{5}\cos 5x+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\sin 2x,\cos 3x,dx}=\frac{1}{2}\int_{{}}^{{}}{2(\sin 2x\cos 3x),dx} $ $ =\frac{1}{2}\int_{{}}^{{}}{(\sin 5x-\sin x),dx}=\frac{1}{2}[ -\frac{\cos 5x}{5}+\cos x ]+c $ $ =\frac{1}{2}[ \cos x-\frac{\cos 5x}{5} ]+c. $
BETA
