Question

Evaluate $\int x\sin 3xdx$

• A. $-\frac{1}{3}x\cos 3x+\frac{1}{9}\sin 3x+C$
• B. $-\frac{1}{3}x\cos 3x+\frac{1}{3}\sin 3x+C$
• C. $\frac{1}{3}x\sin 3x-\frac{1}{3}\cos 3x+C$
• D. None of these

Answer 1

The correct option is A $-\frac{1}{3}x\cos 3x+\frac{1}{9}\sin 3x+C$

$I=\int x\sin 3xdx$

Using Integration by parts,

$I=x\int \sin 3xdx-\int \left(\frac{d}{dx}\right(x)\times \int \sin 3xdx)dx$

$I=x\times -\frac{1}{3}\cos 3x-\int -\frac{1}{3}\cos 3xdx$

$I=-\frac{1}{3}x\cos 3x+\frac{1}{3}\int \cos 3xdx=-\frac{1}{3}x\cos 3x+\frac{1}{9}\sin 3x+C$