Integral Calculus Question 34
Question: If $ \int_{{}}^{{}}{\sin 5x\cos 3x\ dx=-\frac{\cos 8x}{16}}+A $ , then $ A= $
[MP PET 1992]
Options:
A) $ \frac{\sin 2x}{16}+ $ constant
B) $ -\frac{\cos 2x}{4}+ $ constant
C) Constant
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\sin 5x\cos 3x\ dx}=\frac{1}{2}\int_{{}}^{{}}{(\sin 8x+\sin 2x)}dx $ $ =\frac{-\cos 8x}{16}-\frac{\cos 2x}{4}+c $ Equating to the given value, we get $ A=\frac{-\cos 2x}{4}+c $ .
BETA
