Use integration by parts to establish a reduction formula for the integral.
-$\int \sin ^ { n } x d x$
A) $\int \sin ^ { n } x d x = \sin ^ { n - 1 } x \cos x + ( n - 1 ) \int \sin ^ { n - 2 } x d x$
B) $\int \sin ^ { n } x d x = - \frac { 1 } { n } \sin ^ { n - 1 } x \cos x - \frac { n - 1 } { n } \int \sin ^ { n - 1 } x d x$
C) $\int \sin ^ { n } x d x = \sin ^ { n - 1 } x \cos x - ( n - 1 ) \int \cos x \sin ^ { n - 2 } x d x$
D) $\int \sin ^ { n } x d x = - \frac { 1 } { n } \sin ^ { n - 1 } x \cos x + \frac { n - 1 } { n } \int \sin ^ { n - 2 } x d x$

Question

Quizplus · Accepted Answer

The correct formula is derived by applying integration by parts, using $u = \sin^{n-1}x$ and $dv = \sin x dx$, leading to the reduction formula where each integral of $\sin^n x$ is expressed in terms of an integral of $\sin^{n-2}x$, facilitating the process of integration by reducing the power of the sine function step by step.

Use Integration by Parts to Establish a Reduction Formula for the Integral