JEE Main201910 Jan 2019Morning ShiftMathematicsIndefinite IntegrationActual
Let, n ≥ 2 be a natural number and 0 < θ < π 2 . Then ∫ s i n n θ - s i n θ 1 n c o s θ s i n n + 1 θ d θ , is equal to
Options
- An n 2 - 1 1 - 1 s i n n + 1 θ n + 1 n + c
- Bn n 2 + 1 1 - 1 s i n n - 1 θ n + 1 n + c
- Cn n 2 - 1 1 - 1 s i n n - 1 θ n + 1 n + c
- Dn n 2 - 1 1 + 1 s i n n - 1 θ n + 1 n + c
Correct answer
C. n n 2 - 1 1 - 1 s i n n - 1 θ n + 1 n + c
Step-by-step solution
∫ sin n θ - sin θ 1 n cos θ sin n + 1 θ d θ Put, sin θ = t ⇒ cos θ d θ = d t   = ∫ t n - t 1 n d t t n + 1 = ∫ t 1 - 1 t n - 1 1 n t n + 1 d t =   ∫ 1 - 1 t n - 1 1 n t n d t Put 1 - 1 t n - 1 = z ⇒ n - 1 t n d t = d z ⇒ I = 1 n - 1 ∫ z 1 n d z Using ∫ x n d x = x n + 1 n + 1 + c ⇒ I = z 1 n + 1 1 n + 1 n - 1 + c ⇒ I = n 1 - t 1 - n 1 n + 1 n 2 - 1 + c , where c is the constant of integration. ⇒