Question

Question #73167

Answer 1

(sinx -xcosx)/(xsinx+cosx)+C

Explanation:

`int (x^2 dx)/((x sin x +cos x)^2)= int -x/(cos x) d(1/(xsinx+cosx))`

`=(-xsecx)/(xsin x+cos x)+int (sec x + xsecx tanx)/(xsinx+cosx) dx`

But
`(sec x + xsecx tanx)/(xsinx+cosx) =(1/(cos x) + (xsinx)/(cos^2x))/(xsin x+ cosx)`

=`1/cos^2x\cancel (xsinx +cosx)/\cancel(xsinx+cosx)`
=`sec^2x`

then
`int (x^2 dx)/((x sin x +cos x)^2) = (-xsecx)/(xsin x+cos x) + int sec^2x dx`

`= (-xsecx)/(xsin x+cos x)+tanx + C`
`= (xsin^2x + sin xcos x - x)/(cos x(xsin x+cos x))+C`
`=(sinx -xcosx)/(xsinx+cosx)+C`

We can verify by differentiating

`f'(x) = ((xsin x+cos x)(\cancel(cosx)-\cancel(cosx)+xsinx) - (sinx - xcosx)(xcosx + \cancel (sinx) - cancel(sinx)))/(xsinx+cosx)^2`

`=(x^2sin^2x+\cancel(xsinxcosx)-\cancel(xsinxcosx)+x^2cos^2x)/(xsinx+cosx)^2`
= `x^2/(xsinx+cosx)^2`