MHB Integrating $\int x^2\cos\left({\frac{x}{2}}\right)dx$ by parts

[karush]

$\int x^2\cos\left({\frac{x}{2}}\right)dx$
$u={x}^{2}\ dv=\cos{\left(\frac{x}{2}\right)}dx$
$du=2x dx\ v=\int\cos\left({\frac{x}{2}}\right)dx=2\sin\left({\frac{x}{2}}\right)$

Integrat by parts, just seeing if getting started ok

 

[Siron]

$\int x^2\cos\left({\frac{x}{2}}\right)dx$
$u={x}^{2}\ dv=\cos{\left(\frac{x}{2}\right)}dx$
$du=2x dx\ v=\int\cos\left({\frac{x}{2}}\right)dx=2\sin\left({\frac{x}{2}}\right)$

Integrat by parts, just seeing if getting started ok

That is correct! Go on ...

 

[karush]

$$uv-\int v \ du$$

So

$$2x^2\sin\left({\frac{x}{2}}\right)-4\int x \sin\left({\frac{x}{2}}\right)dx$$

Look like another round of parts..

 

[Dethrone]

Yup, you should notice that your original integrand consists of two functions, a polynomial and a trigonometric function. Each time you apply IBP, the degree of your polynomial should decrement by 1, eventually turning into a constant. Then, you will encounter either $\sin\left({\frac{x}{2}}\right)$ or $\cos\left({\frac{x}{2}}\right)$ which you can easily integrate.

 

[karush]

$\int x\sin{\left(\frac{x}{2}\right)}du$
$u=x\ dv=\sin\left({\frac{x}{2}}\right)dx$
$du=dx\ v=\int\sin{\left(\frac{x}{2}\right)}dx=-2\cos{\left(\frac{x}{2}\right)}$
$4\left[-2x\cos\left({\frac{x}{2}}\right)-4\sin\left({\frac{x}{2}}\right)\right]$

 

[karush]

$2x^2\sin\left({\frac{x}{2}}\right)+8x\cos\left({\frac{x}{2}}\right)-16\sin\left({\frac{x}{2}}\right)$
My final answer (I hope)

 

[Prove It]

$2x^2\sin\left({\frac{x}{2}}\right)+8x\cos\left({\frac{x}{2}}\right)-16\sin\left({\frac{x}{2}}\right)$
My final answer (I hope)

+ C

 

[karush]

I seem to forget that to often...