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Integrate(e^sinx(xcos^3x - sinx)/cos^2x)
Integral [e^sinx(xcos^3x-sinx)/cos^2x] dx=integral e^sinx .xcosx dx-integral e^sinx.sinx/cos^2x dx=integral e^sinx.xcosx dx-integrale^sinx.tanxsecx dxlet sinx=t in first integral then on diff w.r. to x we get cosx dx=dt=>integrale^t.sininverset dt-integrale^sinx.tanxsecxdxintegrating by parts Taking sininverset as first function and e^t as second function in first integral and e^sinx as first function and tanxsecx as second function in second integral, we getsininverset.e^t-integral(1/root(1-t^2) .e^t) dt-[e^sinx.secx-integral(e^sinx.cosx.secx )dx]=e^tsininverset-integral(e^t/root(1-t^2))dt-e^sinx.secx+integral(e^sinx)dx=xe^sinx.-integral(e^t/root(1-t^2))dt-e^sinx.secx+integral(e^sinx)dxusing sinx=t and cosxdx=dt, we get=>xe^sinx-integral(e^sinx)dx-e^sinx.secx+integral(e^sinx) dx=xe^sinx-e^sinx.secx +C=(x-secx).e^sinx +CThanks & RegardsRinkoo GuptaAskIITians Faculty
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