2
$\begingroup$

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!

  • 0
    I use ILATE formula to do this integration.I means inverse trig.functions,L stands for logarithmic , A for airthmatic, T for trigonometric, E for exponential functions. using this i can select which function will be first.because in later part of integration when two times integration has been performed it will messy2013-05-10

2 Answers 2

1

$\int \sin^{n}(x) dx = \int \sin^{n-1}(x) \sin(x) dx$ Which one easier to integrate? That will be the dv term. The rest is u. The choice was made thinking about dv, not about u.

0

I'm not really sure there's an exact science to solving integrals. Seriously, how many people would have known to solve $\int \sec xdx$ by multiplying by $\frac{\sec x+\tan x}{\sec x+\tan x}$ without being taught?

All we can do is try to find a way to make things simpler and see if it works. Selecting dv=dx is going to introduce an x into the integral. Usually this is a bad thing, unless you're solving $\int\ln xdx$. Usually, any powers of x make a good choice of u for integration by parts as taking derivatives of it will eventually make it go away.

So dv=dx is probably not a good idea. Selecting $u=\sin x$ would make $dv=\sin^{n-1}xdx$, which looks very similar to the problem you have to begin with. If you could integrate it easily, you probably wouldn't be messing with integration by parts for this problem in the first place. So what else can we choose for dv that we can integrate easily?

  • 1
    There is a systematic procedure for integrating rational functions of trigonometric functions using the Weierstrass substitution (http://en.wikipedia.org/wiki/Weierstrass_substitution). It involves zero guesswork.2012-03-02