The question:
Compute∫x2dx(xsinx+cosx)2
Tried integration by parts. That didn't work.
How do I proceed?
Answer
Observe that, d(xsinx+cosx)dx=xcosx
∫x2dx(xsinx+cosx)2=∫xcosx⋅xcosx(xsinx+cosx)2dx
So, if z=xsinx+cosx,dz=xcosxdx
So, ∫xcosx(xsinx+sinx)2dx=∫dzz2=−1z=−1xsinx+cosx
So, I=xcosx∫xcosx(xsinx+cosx)2dx−∫(d(xcosx)dx∫xcosx(xsinx+cosx)2dx)dx
=−xcosx(xsinx+cosx)+∫(xsinx+cosxcos2x)(1xsinx+cosx)dx
=−xcosx(xsinx+cosx)+∫sec2xdx
=−xcosx(xsinx+cosx)+tanx+C where C is an arbitrary constant of indefinite integral
Another form will be sinx−xcosxxsinx+cosx+C
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