Question

The value of the integral $\int \sin \left(60x\right){\sin }^{58}xdx$ is
(where ${}^{\prime }{C}^{\prime }$ is the constant of integration)

• A. $\frac{\sin \left(59x\right){\sin }^{58}x}{58}+C$
• B. $\frac{\sin \left(58x\right){\sin }^{59}x}{60}+C$
• C. $\frac{\sin \left(60x\right){\sin }^{59}x}{59}+C$
• D. $\frac{\sin \left(59x\right){\sin }^{59}x}{59}+C$

Answer 1

The correct option is D $\frac{\sin \left(59x\right){\sin }^{59}x}{59}+C$

$I=\int \sin \left(60x\right){\sin }^{58}xdx$
$=\int \sin (59x+x){\sin }^{58}xdx$
$=\int \left(\sin \right(59x)\cos x+\cos (59x\left)\sin x\right){\sin }^{58}xdx$
$=\int \left[\sin \right(59x){\sin }^{58}x\cos x+{\sin }^{59}x\cos (59x\left)\right]dx$

Integrate $\sin \left(59x\right){\sin }^{58}x\cos x$ by parts, we get

$=\frac{\sin \left(59x\right){\sin }^{59}x}{59}-\int \cos \left(59x\right){\sin }^{59}xdx+\int \cos \left(59x\right){\sin }^{59}xdx$
$=\frac{\sin \left(59x\right){\sin }^{59}x}{59}+C$