A) \[x+\sin 2x+c\]
B) \[3x+\sin 2x+c\]
C) \[3x+{{\sin }^{2}}x+c\]
D) None of these
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{\sin 3x}{\sin x}\,dx}=\int_{{}}^{{}}{\frac{3\sin x-4{{\sin }^{3}}x}{\sin x}\,dx}\] \[\int_{{}}^{{}}{3\,dx}-4\int_{{}}^{{}}{{{\sin }^{2}}x\,dx}=3x-2\int_{{}}^{{}}{(1-\cos 2x)\,dx+c}\] \[=3x-2x+\sin 2x+c=x+\sin 2x+c.\]You need to login to perform this action.
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