We have,
∫exsinxdx
Using product rule and we get,
∫I.IIdx=I∫IIdx−∫(dIdx∫IIdx)dx
And we get,
I=∫exsinxdx=sinx∫exdx−∫(ddxsinx∫exdx)dx
I=sinxex−∫excosxdx
I=sinxex−[cosx∫exdx−∫ddxcosx∫exdx]
I=sinxex−[excosx−∫ex(−sinx)]
I=exsinx−excosx−∫exsinx
I=exsinx−excosx−I
I+I=exsinx−excosx
2I=exsinx−excosx
I=12ex(sinx−cosx)
Hence, this is the
answer.