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!

  • $\begingroup$ The best way to understand why it was done the way it was done is to try to do it one of the other ways you suggest and see why it doesn't get you anywhere. $\endgroup$ Mar 2, 2012 at 2:56
  • $\begingroup$ Hmm... yes, but then I would like an explanation as to why the other methods wouldn't work. $\endgroup$ Mar 2, 2012 at 2:58
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    $\begingroup$ Note that given the integrand $fg$, when choosing what's what, you're essentially trying to decide which of $f' \int g$ and $g'\int f$ is easier to form and integrate. So for example, given $x^2 \sin x$ the choices are $(2x)(-\cos x)$ and $({x^3\over 3})(\cos x)$. Which of the two would be easier to integrate? (Of course you can also decide to do something sneaky like with $\ln x=1\cdot\ln x$.) $\endgroup$ Mar 2, 2012 at 3:06
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    $\begingroup$ Because the integral of $sin(x)$ is $cos(x)$. Once that pops up in the new integrand the simple substitution $u=sin(x)$ will handle the remaining powers of $sin(x)$. Does it make sense? $\endgroup$
    – Patrick
    Mar 2, 2012 at 3:16
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    $\begingroup$ spryn0724, there's nothing stopping you from choosing $u = \sin^n x$ and $dv = dx$, but you'd get a different formula by doing it. $\endgroup$ Mar 2, 2012 at 3:17

2 Answers 2


$$\int \sin^{n}(x) dx = \int \sin^{n-1}(x) \sin(x) dx$$ 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.


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?


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