Integral Calculus Question 442
Question: $ \int_{{}}^{{}}{\frac{\sin 2x}{\sin 5x\sin 3x}}\ dx= $
Options:
A) $ \log \sin 3x-\log \sin 5x+c $
B) $ \frac{1}{3}\log \sin 3x+\frac{1}{5}\log \sin 5x+c $
C) $ \frac{1}{3}\log \sin 3x-\frac{1}{5}\log \sin 5x+c $
D) $ 3\log \sin 3x-5\log \sin 5x+c $
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Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{\sin 2x}{\sin 5x\sin 3x},dx}=\int_{{}}^{{}}{\frac{\sin (5x-3x)}{\sin 5x\sin 3x},dx} $ $ =\int_{{}}^{{}}{\frac{\sin 5x\cos 3x-\cos 5x\sin 3x}{\sin 5x\sin 3x},dx} $ $ =\frac{1}{3}\log \sin 3x-\frac{1}{5}\log \sin 5x+c. $
BETA
