Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
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
	$Ω(n):= ∫_0 ^( (π/4)) ((ln(cot(x)).sin^(n−1) (2x))/((sin^n (x)+cos^n (x))^2 ))dx =2^(n−2) ∫_0 ^( (π/4)) ((ln(cot(x)).sin^(n−1) (x).cos^(n−1) (x))/(sin^(2n) (x)(1+cot^n (x))^2 ))dx =(2^(n−1) /n^2 )∫_0 ^( (π/4)) ((ln(cot^n (x)).sin^(n−1) (x)(n).cos^(n−1) (x))/(sin^2 (x).sin^(n−1) (x)sin^(n−1) (x)(1+cot^n (x))^2 ))dx =(2^(n−1) /n^2 )∫_0 ^( (π/4)) ((ln(cot^n (x)).ncot^(n−1) (x))/(sin^2 (x)(1+cot^n (x))^2 ))dx =^(cot^n (x)=y) (2^(n−1) /n^2 )∫_1 ^( ∞) ((ln(y)dy)/((1+y)^2 )) =(2^(n−1) /n^2 ) {[((−1)/(1+y))ln(y)]_1 ^( ∞) +∫_1 ^^∞ (1/(y(1+y)))dy dy} =(2^(n−1) /n^2 ){ln((y/(1+y)))}_1 ^∞ =(2^(n−1) /n^2 )ln(2) n:=π^e +1 ........ ....Ω =((2^( π^( e) ) . ln(2))/((π^( e) +1)^2 )) .....$
	$Ω (n) := \int_{0}^{\frac{π}{4}} \frac{l n (c o t (x)) . {s i n}^{n - 1} (2 x)}{{({s i n}^{n} (x) + {c o s}^{n} (x))}^{2}} d x$ $= 2^{n - 2} \int_{0}^{\frac{π}{4}} \frac{l n (c o t (x)) . {s i n}^{n - 1} (x) . {c o s}^{n - 1} (x)}{{s i n}^{2 n} (x) {(1 + {c o t}^{n} (x))}^{2}} d x$ $= \frac{2^{n - 1}}{n^{2}} \int_{0}^{\frac{π}{4}} \frac{l n ({c o t}^{n} (x)) . {s i n}^{n - 1} (x) (n) . {c o s}^{n - 1} (x)}{{s i n}^{2} (x) . {s i n}^{n - 1} (x) {s i n}^{n - 1} (x) {(1 + {c o t}^{n} (x))}^{2}} d x$ $= \frac{2^{n - 1}}{n^{2}} \int_{0}^{\frac{π}{4}} \frac{l n ({c o t}^{n} (x)) . {n c o t}^{n - 1} (x)}{{s i n}^{2} (x) {(1 + {c o t}^{n} (x))}^{2}} d x$ $\overset{{c o t}^{n} (x) = y}{=} \frac{2^{n - 1}}{n^{2}} \int_{1}^{\infty} \frac{l n (y) d y}{{(1 + y)}^{2}}$ $= \frac{2^{n - 1}}{n^{2}} {{[\frac{- 1}{1 + y} l n (y)]}_{1}^{\infty} + \int_{1}^{^{\infty}} \frac{1}{y (1 + y)} d y$ $d y}$ $= \frac{2^{n - 1}}{n^{2}} {l n (\frac{y}{1 + y})}_{1}^{\infty} = \frac{2^{n - 1}}{n^{2}} l n (2)$ $n := π^{e} + 1 . . . . . . . .$ $. . . . Ω = \frac{2^{π^{e}} . l n (2)}{{(π^{e} + 1)}^{2}} . . . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum