I am looking at an example of integration-by-parts in my Calculus book, and there is one thing that I do not understand:
Prove the reduction formula: $\int \sin^n x \ dx = -\frac{1}{n} \cdot \cos x \cdot \sin^{n - 1} x + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$
This problem clearly requires the use of integration-by-parts. I am very comfortable with integration-by-parts, but, in this example, I don't understand why they chose $u$ and $dv$ the way they did:
$\int \sin^n x \ dx = \int \underbrace{\sin^{n - 1}x}_u \cdot \underbrace{\sin x \ dx}_{dv}$
$u = \sin^{n - 1} x$
$du = -(n - 1) \cdot \sin^{n - 2} x \cdot \cos x \ dx$
$v = -\cos x$
$dv = \sin x \ dx$
... and the rest of the problem is solved ...
In previous examples, such as:
Find $\displaystyle\int \ln x \ dx$.
... I was told to let u
be $\ln x$ in the equation, and, of course, $dv$ would end up being everything else, namely $1 \cdot dx$.
Why, then, did they decide to split up $\sin^n x$ into two terms and then let $dv$ be $\sin{x} \ dx$ rather than the understood 1?
Also, why was $\sin^{n - 1} x$ chosen for u
rather than $\sin x$?
Thank you for your time!