Evaluate-0-pi-4-ln-tan-x-sin-pi-e-2x-sin-pi-e-x-cos-pi-e-x-2-dx-

Question Number 143086 by mnjuly1970 last updated on 09/Jun/21

E v a l u a t e ::

Ω := \int_{0}^{\frac{π}{4}} \frac{l n (t a n (x)) . {s i n}^{π^{e}} (2 x)}{{({s i n}^{π^{e}} (x) + {c o s}^{π^{e}} (x))}^{2}} d x

Answered by alexperez2703a last updated on 10/Jun/21

E v a l u a t e ::

\int_{0}^{3} (4 - x) (3 - x) d x =

Answered by mnjuly1970 last updated on 17/Jun/21

Ω (n) := \int_{0}^{\frac{π}{4}} \frac{l n (c o t (x)) . {s i n}^{n - 1} (2 x)}{{({s i n}^{n} (x) + {c o s}^{n} (x))}^{2}} d x

= 2^{n - 2} \int_{0}^{\frac{π}{4}} \frac{l n (c o t (x)) . {s i n}^{n - 1} (x) . {c o s}^{n - 1} (x)}{{s i n}^{2 n} (x) {(1 + {c o t}^{n} (x))}^{2}} d x

= \frac{2^{n - 1}}{n^{2}} \int_{0}^{\frac{π}{4}} \frac{l n ({c o t}^{n} (x)) . {s i n}^{n - 1} (x) (n) . {c o s}^{n - 1} (x)}{{s i n}^{2} (x) . {s i n}^{n - 1} (x) {s i n}^{n - 1} (x) {(1 + {c o t}^{n} (x))}^{2}} d x

= \frac{2^{n - 1}}{n^{2}} \int_{0}^{\frac{π}{4}} \frac{l n ({c o t}^{n} (x)) . {n c o t}^{n - 1} (x)}{{s i n}^{2} (x) {(1 + {c o t}^{n} (x))}^{2}} d x

\overset{{c o t}^{n} (x) = y}{=} \frac{2^{n - 1}}{n^{2}} \int_{1}^{\infty} \frac{l n (y) d y}{{(1 + y)}^{2}}

= \frac{2^{n - 1}}{n^{2}} {{[\frac{- 1}{1 + y} l n (y)]}_{1}^{\infty} + \int_{1}^{^{\infty}} \frac{1}{y (1 + y)} d y

d y}

= \frac{2^{n - 1}}{n^{2}} {l n (\frac{y}{1 + y})}_{1}^{\infty} = \frac{2^{n - 1}}{n^{2}} l n (2)

n := π^{e} + 1 \dots \dots . .

\dots . Ω = \frac{2^{π^{e}} . l n (2)}{{(π^{e} + 1)}^{2}} \dots . .