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I am trying to compute the following integral:

$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$

I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:

Let $u = \cos \frac{x}{2}$ and $\mathrm{d}u = \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ and $\mathrm{d}v = x^2$ and $v= \frac{x^3}{3}$.

\begin{align} &\int x^2 \cos \frac{x}{2} \\ &\cos \frac{x}{2} \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x \end{align}

So now I integrate $\int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ to get:

\begin{align} \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \cdot \frac{x^4}{12} - \int \frac{x^4}{12} \cdot \frac{\cos x}{2} \end{align}

Now, this is where I get stuck. I know if I continue, I will end up with $\frac{-\sin\left(\dfrac{x}{2}\right)}{2}$ again when I integrate $\cos \frac{x}{2}$. So, where do I go from here?

Thanks!

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4 Answers 4

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$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$

Let $u = x^2$ and $\mathrm{d}u = 2x\,dx$ and let $dv = \cos\frac x2\,dx \implies v = 2 \sin \frac x2 $.

$$ 2x^2 \sin \frac x2 - \int 2x\cdot 2\sin \frac x2\,dx $$

You'll only need to do integration by parts one additional time. Let me know if you get stuck after that.

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  • $\begingroup$ I am having trouble simplifying $2x^2 \sin (x/2) + 8x \cos (x/2) - 16 \sin (x/2)$. I factored out a $2$ but what then? $\endgroup$
    – Jeel Shah
    Aug 16, 2013 at 23:17
  • $\begingroup$ You'll have $2x^2 \sin (x/2) + 8x \cos (x/2) - 16 \sin (x/2) + C = (2x^2 - 16)\sin(x/2) + 8x \cos(x/2) + C$. No need to simplify further. If you've access to the solution, and it differs, there may be a half-angle identity you can use. $\endgroup$
    – amWhy
    Aug 16, 2013 at 23:22
  • $\begingroup$ Ahh! Got it! Thanks! $\endgroup$
    – Jeel Shah
    Aug 16, 2013 at 23:26
  • $\begingroup$ You're welcome! $\endgroup$
    – amWhy
    Aug 16, 2013 at 23:27
  • $\begingroup$ @amWhy: great feedback +1 $\endgroup$
    – Amzoti
    Aug 17, 2013 at 0:11
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HINT

Go in the opposite direction. Let $u = x^2$, $dv = \cos \frac{x}{2}$.

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  • $\begingroup$ You are really fast in typing, I was writing this :-) $\endgroup$
    – SomeOne
    Aug 16, 2013 at 22:49
  • $\begingroup$ You mean certainly $v'= \cos \frac{x}{2}$. $\endgroup$
    – user63181
    Aug 16, 2013 at 22:59
  • $\begingroup$ @Sami: I adopted the same notation the OP used in his OP. $\endgroup$
    – davidlowryduda
    Aug 17, 2013 at 4:13
  • $\begingroup$ This is not a simple notation but a differentiation's operator so either we write $dv=\cos \frac{x}{2}dx$ or we write $v'=\cos \frac{x}{2}$ isn't it? $\endgroup$
    – user63181
    Aug 17, 2013 at 8:43
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Hint: Switch the functions you originally set equal to $u$ and $dv$.

General Strategy: Assume you're trying to compute $\int p(x) f(x) dx$ where $p(x) = a_0 + a_1 x + \ldots + a_nx^n$ is a polynomial and $f(x)$ is a function that you know how to integrate $n$ times. (For example, in your problem $p(x) = x^2$, $f(x) = \cos x$, and we note that it's easy to integrate $\cos x$ twice.) Then you can just keep integrating by parts ($n$ times), setting $u = p(x)$ (or $p^{(k)}(x)$) and $dv = \lbrack\text{whatever is leftover}\rbrack dx$ to eventually arrive at $$ \int p(x) f(x) dx = \lbrack\text{a bunch of terms not involving integrals}\rbrack + \int F(x) dx $$ where $F(x)$ is a function such that $F^{(n)}(x) = f(x)$.

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let S be the integrate... That is, S x"2cosx/2 dx........let u=x"2 and dv=cosx/2dx. so that, du=2x and v=2sinx/2 dx. Therefore, S x"2cosx/2 dx= x"2(2sinx/2)-S 2sinx/2(2x) =2x"2sinx/2-S 2x.2sinx/2 dx = 2x"2sinx/2-4 S xcosx/2 dx. Apply another integration. S xcosx/2 dx. let u=x and dv=cosx/2dx. so that du=(1)dx=dx and v=2sinx/2 dx. Hence, S xcosx/2 dx=x(2cosx/2)-S 2sinx/2 dx. combining all formulas we get. S x"2cosx/2dx=2x"2sin(x/2)-2(4xcos(x/2)-S 2sin(x/2) dx). That is, S x"2sin(x/2)+8xcos(x/2)-16sin(x/2) dx = 2x"2sin(x/2)+8xcos(x/2)-16sin(x/2)+C. SO, (2x"2-16)sin(x/2)+8xcos(x/2)+C. I hope this help. By O'john

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