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How to create a reduction formula for the integral$$\int x \cos^n x \;\mathrm{d}x$$

I have tried everything i could think of, but I'm not even close to solving it. Really need help.

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    What about parts? That's a start. – 2017-02-20

4 Answers 4

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Let $I_n$ be the integral

$$I_n=\int x\cos^n(x)\,dx$$

Then, we have

$$\begin{align} I_{n+2}&=\int x\cos^2(x)\cos^n(x)\,dx\\\\ &=I_n-\int x\sin^2(x)\cos^n(x)\,dx\tag 1 \end{align}$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=x\sin(x)$ and $v=-\frac{\cos^{n+1}(x)}{n+1}$ yields

$$\begin{align} I_{n+2}&=I_n-\left(x\sin(x)\left(-\frac{\cos^{n+1}(x)}{n+1}\right)+\frac{1}{n+1}\int (x\cos(x)+\sin(x))\cos^{n+1}(x)\,dx\right)\\\\ &=I_n+\frac{x\sin(x)\cos^{n+1}(x)}{n+1}-\frac{1}{n+1}I_{n+2}+\frac{\cos^{n+2}(x)}{(n+1)(n+2)}\\\\ &=\frac{n+1}{n+2}I_n+\frac{x\sin(x)\cos^{n+1}(x)}{n+2}+\frac{\cos^{n+2}(x)}{(n+2)^2} \end{align}$$

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    This is a nice, simple and elegant approach, for sure ! – 2017-02-20
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    Please i don't understand how [∫x(cos²x)(cosx)^n dx] changed to the right hand side of (1) – 2017-02-20
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    I just got it, thank you @Dr. MV and everyone else. – 2017-02-20
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    You're welcome. My pleasure. -Mark – 2017-02-20
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    And please feel free to up vote and best vote as you see fit of course. – 2017-02-20
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    @claudeleibovici Thank you my friend, as always. – 2017-02-20
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Hint:

A reduction formula for $$\int \cos^n x \;\mathrm{d}x$$

is $$\int \frac{\cos^{n-1}(x)\sin(x)}{n}$$

The additional 'x' term would be an application of integration by parts.

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    First, that is not the reduction formula for $\int \cos^n(x)\,dx$. Second, the statement regarding "The addition '$x$' term ..." is the whole source of the difficulty. Have you tried pursuing a solution? – 2017-02-20
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By parts, integrating $\cos x$, $$I_n:=\int x \cos^n x \,dx=x\sin x\cos^{n-1}x-\int\sin x\,(x\cos^{n-1}x)'\,dx \\=x\sin x\cos^{n-1}x-\int\sin x\, \cos^{n-1}x\,dx+(n-1)\int x\sin^2x\,\cos^{n-2}x\,dx \\=x\sin x\cos^{n-1}x+\frac1n\cos^nx+(n-1)\int x\,\cos^{n-2}x\,dx-(n-1)\int x\,\cos^nx\,dx .$$

So, we get the recurrence relation

$$nI_n=x\sin x\cos^{n-1}x+\frac1n\cos^nx+(n-1)I_{n-2}.$$

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You can actually evaluate this integral for general $n$ in terms of hypergeometric functions.

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    That is possible, but how does that answer the actual question? Do you think OP is aware of a recurrence relation for the hypergeometric functions? – 2017-02-20