[SOLVED] evaulating the integral- integration by parts

**genlovesmusic09** · December 6th 2009, 11:50 AM

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

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

**Chris L T521** · December 6th 2009, 11:53 AM

Originally Posted by genlovesmusic09

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

I tried:
u=xsinx, du=xcox+sinx
v=sinx, dv=cosx
$(x\sin x)(\sin x) - \int {(xcoxs + \sin x)(\sin x)dx} $
$(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 $\tfrac{1}{2}\int_0^{\infty} x\sin\!\left(2x\right)\,dx$

**Defunkt** · December 6th 2009, 12:56 PM

Originally Posted by genlovesmusic09

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

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

Another way:

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

We get:

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

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

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