Let n≥2 be a natural number and 0<θ<π2. Then, ∫(sinnθ−sinθ)1ncosθsinn+1θdθ is equal to (where C is a constant of integration)

Let I=∫(sinnθ−sinθ)1/ncosθsinn+1θdθ Put sinθ=t⇒cosθdθ=dt ∴  I=∫(tn=t)1/ntn+1dt =∫[tn(1−ttn)]1/ntn+1 =∫t(1−1/tn−1)1/ntn+1dt=∫(1−1/tn−1)1/ntndt Put 1=1tn−1=u Or 1−t−(n−1)=u⇒(n−1)tndt=du ⇒  dttn=dun−1 ⇒t=∫u1/ndun−1=u1n+1(n−1)(1n+1)+C =n(1−1tn−1)n+1n(n−1)(n+1)+C =n(1−1sinn−1θ)n+1n(n−1)(n+1)+C =n(1−1sinn−1θ)n+1nn2−1+C [∵  u=1−1tn−1  and  k=sinθ]