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

Question

Quizplus · Accepted Answer

We use integration by parts with $u = \cos^{n-1}x$ and $dv = \cos x dx$, then we get $$du = (n-1) \cos^{n-2}x (-\sin x) dx$$ and $$v = \sin x.$$ Therefore, \begin{align*}
\int \cos^n x dx &= uv - \int v du \
&= \cos^{n-1}x \sin x - \int (n-1)\cos^{n-2}x(-\sin x) \sin x dx \
&= \cos^{n-1}x \sin x + (n-1)\int \cos^{n-2}x \sin^2 x dx \
&= \cos^{n-1}x \sin x + (n-1)\int \cos^{n-2}x (1-\cos^2 x)dx \
&= \cos^{n-1}x \sin x + (n-1)\int \cos^{n-2}x dx - (n-1)\int \cos^{n}x dx.
\end{align*} Rearranging terms, we get $$\int \cos^n x dx = \frac{1}{n}\cos^{n-1}x \sin x + \frac{n-1}{n} \int \cos^{n-2}x dx,$$ which is the desired reduction formula. Thus the best choice is D.

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