Given, x=(2n+1)2π Now, ∫(1+sinx)4cos3xdx=∫(1+sinx)4cosx(1−sin2x)dx On putting 1+sinx=tcosxdx=dt∫t41−(t−1)2dt=∫t4−t2+2tdt=∫(t32−t21)dt=−t21+t1+c=t2t−1+cdt=(1+sinx)2sinx+c The given options are wrong in this question. The correct answer is (1+sinx)2sinx+C