We have,
∫xcosx−sinxxsinxdx
⇒∫(xcosxxsinx−sinxxsinx)dx
⇒∫(cotx−x−1)dx
⇒∫cotxdx−∫1xdx∴∫1xdx=logx
⇒logsinx−logx+C
Hence, this is the answer.
Evaluate : ∫x3(x−1)(x2+1)dx.
OR
Evaluate ∫sin x−x cos xx(x+sin x) dx.