Question

integration of tanx/1-sinx dx

Answer 1

∫tanx1−sinxdx=−∫tanxsinx−1dx=−∫sinxcosx(sinx−1)=−∫cosx(sinx(sinx−1)2(sinx+1))dxTake u=sinxdudx=cosxi.e.,=−∫u(u−1)2(u+1)duTaking partial fraction of u(u−1)2(u+1)u(u−1)2(u+1)=au−1+b(u−1)2+cu+1u(u−1)2(u+1)=14(u−1)+12(u−1)2−14(u+1)Thus,−∫u(u−1)2(u+1)du=−[∫14(u−1)du+∫12(u−1)2du−∫14(u+1)du]=−[14ln|u−1|−12(u−1)−14ln|u+1|]=−14ln|sinx−1|+12(sinx−1)+14ln|sinx+1|+C