Wednesday, 22 May 2019

calculus - Evaluating this integral smallintfracx2dx(xsinx+cosx)2




The question:




Computex2dx(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=xcosxxcosx(xsinx+cosx)2dx



So, if z=xsinx+cosx,dz=xcosxdx



So, xcosx(xsinx+sinx)2dx=dzz2=1z=1xsinx+cosx




So, I=xcosxxcosx(xsinx+cosx)2dx(d(xcosx)dxxcosx(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 sinxxcosxxsinx+cosx+C



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