Integral Calculus Question 298
Question: $ \int_{{}}^{{}}{\frac{\sin 3x}{\sin x}\ dx=} $
Options:
A) $ x+\sin 2x+c $
B) $ 3x+\sin 2x+c $
C) $ 3x+{{\sin }^{2}}x+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{\sin 3x}{\sin x},dx}=\int_{{}}^{{}}{\frac{3\sin x-4{{\sin }^{3}}x}{\sin x},dx} $ $ \int_{{}}^{{}}{3,dx}-4\int_{{}}^{{}}{{{\sin }^{2}}x,dx}=3x-2\int_{{}}^{{}}{(1-\cos 2x),dx+c} $ $ =3x-2x+\sin 2x+c=x+\sin 2x+c. $
BETA
