Question

( 3sinx-2).cosx)/5-cos2x-4sinx

Answer 1

=∫(3sinx−2)cosx5−cos2x−4sinxdx=∫(3sinx−2)cosxsin2x−4sinx+4dx=∫(3sinx−2)cosx(sinx−2)2dx

Let sinx=t
So, cosxdx=dt

∴I=∫(3t−2)(t−2)2dt=3∫tdt(t−2)2−2∫dt(t−2)2=3[t(1−3(t−2)3)−∫dt−3(t−2)3]−2(1−3(t−2)3)+C

=−t(t−2)3+13[1−4(t−2)4]+23(t−2)3+C=1(t−2)3(−t+23)−112(t−2)4+C=−1(t−2)2−43(t−2)3−112(t−2)4+C

=−1(sinx−2)2−43(sinx−2)3−112(sinx−2)4 +C