Question Number 143086 by mnjuly1970 last updated on 09/Jun/21
$$ \\ $$$$\:\:{Evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Omega:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({tan}\left({x}\right)\right).{sin}^{\pi^{{e}} } \left(\mathrm{2}{x}\right)}{\left({sin}^{\pi^{{e}} } \left({x}\right)+{cos}^{\pi^{{e}} } \left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$ \\ $$
Answered by alexperez2703a last updated on 10/Jun/21
$$ \\ $$$$\:\:{Evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{3}} \left(\mathrm{4}−{x}\right)\left(\mathrm{3}−{x}\right){dx}= \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
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 )) .....](https://www.tinkutara.com/question/Q143752.png)
$$\:\:\:\:\:\Omega\left({n}\right):=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cot}\left({x}\right)\right).{sin}^{{n}−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\left({sin}^{{n}} \left({x}\right)+{cos}^{{n}} \left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$\:=\mathrm{2}^{{n}−\mathrm{2}} \int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cot}\left({x}\right)\right).{sin}^{{n}−\mathrm{1}} \left({x}\right).{cos}^{{n}−\mathrm{1}} \left({x}\right)}{{sin}^{\mathrm{2}{n}} \left({x}\right)\left(\mathrm{1}+{cot}^{{n}} \left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$\:=\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cot}^{{n}} \left({x}\right)\right).{sin}^{{n}−\mathrm{1}} \left({x}\right)\left({n}\right).{cos}^{{n}−\mathrm{1}} \left({x}\right)}{{sin}^{\mathrm{2}} \left({x}\right).{sin}^{{n}−\mathrm{1}} \left({x}\right){sin}^{{n}−\mathrm{1}} \left({x}\right)\left(\mathrm{1}+{cot}^{{n}} \left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:=\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cot}^{{n}} \left({x}\right)\right).{ncot}^{{n}−\mathrm{1}} \left({x}\right)}{{sin}^{\mathrm{2}} \left({x}\right)\left(\mathrm{1}+{cot}^{{n}} \left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\overset{{cot}^{{n}} \left({x}\right)={y}} {=}\:\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left({y}\right){dy}}{\left(\mathrm{1}+{y}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:\left\{\left[\frac{−\mathrm{1}}{\mathrm{1}+{y}}{ln}\left({y}\right)\right]_{\mathrm{1}} ^{\:\infty} +\int_{\mathrm{1}} ^{\:^{\infty} } \frac{\mathrm{1}}{{y}\left(\mathrm{1}+{y}\right)}{dy}\right. \\ $$$$\left.{dy}\right\} \\ $$$$=\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\left\{{ln}\left(\frac{{y}}{\mathrm{1}+{y}}\right)\right\}_{\mathrm{1}} ^{\infty} =\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }{ln}\left(\mathrm{2}\right) \\ $$$$\:\:{n}:=\pi^{{e}} +\mathrm{1}\:…….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\Omega\:=\frac{\mathrm{2}^{\:\pi^{\:{e}} } .\:{ln}\left(\mathrm{2}\right)}{\left(\pi^{\:{e}} +\mathrm{1}\right)^{\mathrm{2}} }\:….. \\ $$$$\: \\ $$