Question-120727

Question Number 120727 by 77731 last updated on 02/Nov/20

Answered by TANMAY PANACEA last updated on 02/Nov/20

N_{r} = 2 + 2 c o s x - 2 c o s n x - c o s (n - 1) x - c o s (n + 1) x

= 2 + 2 c o s x - 2 c o s n x - 2 c o s n x c o s x

= 2 (1 + c o s x) - 2 c o s n x (1 + c o s x)

= 2 (1 + c o s x) (1 - c o s n x)

\frac{N_{r}}{D_{r}} = \frac{2 (1 + c o s x) (1 - c o s n x)}{1 - c o s 2 x} = \frac{2 (1 + c o s x) (1 - c o s n x)}{2 {s i n}^{2} x}

I_{n} - I_{n - 2} = \int_{0}^{π} (\frac{1 + c o s x}{{s i n}^{2} x}) ({1 - c o s n x - 1 + c o s (n - 2) x} d x

= \int_{0}^{π} (\frac{1 + c o s x}{{s i n}^{2} x}) \times 2 s i n (n - 1) x s i n x

= \int_{0}^{π} \frac{1 + c o s x}{s i n x} \times s i n (n - 1) x d x

= \int_{0}^{π} \frac{2 {c o s}^{2} \frac{x}{2}}{2 s i n \frac{x}{2} c o s \frac{x}{2}} \times s i n (n - 1) x d x

= \int_{0}^{π} c o t \frac{x}{2} \times s i n (n - 1) x d x

I_{n} - I_{n - 2} = \int_{0}^{π} c o t \frac{x}{2} s i n (n - 1) x d x

w a i t \dots