Consider the integral I_n=∫0(π)/4(sin2n−1x)/(sin x)dx, then the value of I_20-I_19 is

Question

MARKS Content Team · Accepted Answer

Correct answer: (B) (-1)/19. I_20 −I_19 = ∫ 0 (π)/4 (sin 39x −sin 37x)/(sin x) dx ⇒I_20 −I_19 = ∫ 0 (π)/4 (2sin x cos 38x)/(sin x) dx =sin38x/19 (π)/4 0 =(sin 38π)/4 19 =(-1)/19

Consider the integral $I_{n} = \int_{0}^{\frac{π}{4}} \frac{sin (2 n - 1) x}{sin x} d x,$ then the value of $I_{20} - I_{19}$ is

Solution

More Definite Integration Questions

Consider the integral In=∫0π4sin2n−1xsin xdx, then the value of I20-I19 is

Solution

More Definite Integration Questions

Consider the integral $I_{n} = \int_{0}^{\frac{π}{4}} \frac{sin (2 n - 1) x}{sin x} d x,$ then the value of $I_{20} - I_{19}$ is