# Thread: [SOLVED] evaulating the integral- integration by parts

1. ## [SOLVED] evaulating the integral- integration by parts

$\displaystyle \int_0^\infty {x\sin (x)\cos (x)dx}$

I tried:
u=xsinx, du=xcox+sinx
v=sinx, dv=cosx
$\displaystyle (x\sin x)(\sin x) - \int {(xcoxs + \sin x)(\sin x)dx}$
$\displaystyle (x\sin x)(\sin x) - \int {x\cos x\sin x + {{\sin }^2}xdx}$
but then I got stuck on integrating...

2. Originally Posted by genlovesmusic09
$\displaystyle \int_0^\infty {x\sin (x)\cos (x)dx}$

I tried:
u=xsinx, du=xcox+sinx
v=sinx, dv=cosx
$\displaystyle (x\sin x)(\sin x) - \int {(xcoxs + \sin x)(\sin x)dx}$
$\displaystyle (x\sin x)(\sin x) - \int {x\cos x\sin x + {{\sin }^2}xdx}$
but then I got stuck on integrating...
Recall that 2sinxcosx = sin(2x).

So I would recommend that you rewrite the integral as $\displaystyle \tfrac{1}{2}\int_0^{\infty} x\sin\!\left(2x\right)\,dx$

3. Originally Posted by genlovesmusic09
$\displaystyle \int_0^\infty {x\sin (x)\cos (x)dx}$

I tried:
u=xsinx, du=xcox+sinx
v=sinx, dv=cosx
$\displaystyle (x\sin x)(\sin x) - \int {(xcoxs + \sin x)(\sin x)dx}$
$\displaystyle (x\sin x)(\sin x) - \int {x\cos x\sin x + {{\sin }^2}xdx}$
but then I got stuck on integrating...
Another way:

Since $\displaystyle \int xsin(x)cos(x) = xsin^2(x) - \int xsin(x)cos(x) - \int sin^2(x)$,

We get:

$\displaystyle 2 \int xsin(x)cos(x) = xsin^2(x) - \int sin^2(x)$

From here it should be easy (find the RHS and divide it by 2 to get the answer).