Question

Evaluate : $\int \frac{sinx+4sin3x+6sin5x+3sin7x}{sin2x+3sin4x+3sin6x}dx$

• A. $-2sinx+c$
• B. $2sinx+c$
• C. $-2cosx+c$
• D. $2cosx+c$

Answer 1

Answer: (B)

$2sinx+c$

Given $\int \frac{sinx+4sin3x+6sin5x+3sin7x}{sin2x+3sin4x+3sin6x}dx$

$=\int \frac{(sinx+sin3x)+3(sin3x+sin5x)+3(sin5x+sin7x)}{sin2x+3sin4x+3sin6x}dx$

$\int \frac{(2sin2xcosx+6sin4xcosx+6sin6xcosx)}{sin2x+3sin4x+3sin6x}dx$ $[\because \sin A+\sin B=2\sin (\frac{A+B}{2}\left)\cos \right(\frac{A-B}{2}\left)\right]$

$=\int \frac{2cosx(sin2x+3sin4x+3sin6x)}{sin2x+3sin4x+3sin6x}dx=2\int \cos xdx$ $=2sinx+c$