Question Number 171560 by mnjuly1970 last updated on 17/Jun/22
$$ \\ $$$$\:\:\:\:\:\mathrm{Nice}\:\:\:\mathrm{Integral} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\:{tan}\left({x}\right)}{\left(\:{cos}^{\:\mathrm{2}} \left({x}\right)\:\:+\:\mathrm{2}{sin}^{\:\mathrm{2}} \left({x}\right)\right)}{dx}\:= \\ $$
Commented by infinityaction last updated on 17/Jun/22
![Ω = ∫_0 ^(π/4) ((tan x sec^2 x)/(1+2tan^2 x))dx tan x = p → sec^2 x dx = dp Ω = ∫_0 ^1 (p/(1+2p^2 ))dp 1+2p^2 = t → 4pdp = dt Ω = (1/4)∫_1 ^3 (dt/(t )) ⇒ (1/4)[log t]_1 ^3 Ω = (1/4)log 3](https://www.tinkutara.com/question/Q171561.png)
$$\:\:\:\:\:\:\Omega\:\:\:=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{tan}\:{x}\:\mathrm{sec}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{2tan}\:^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:\:\mathrm{tan}\:{x}\:=\:{p}\:\rightarrow\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}\:\:=\:{dp} \\ $$$$\:\:\:\:\:\:\:\Omega\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{p}}{\mathrm{1}+\mathrm{2}{p}^{\mathrm{2}} }{dp} \\ $$$$\:\:\:\:\:\:\:\mathrm{1}+\mathrm{2}{p}^{\mathrm{2}} \:\:=\:\:{t}\:\:\:\rightarrow\:\:\:\mathrm{4}{pdp}\:=\:{dt} \\ $$$$\:\:\:\:\:\:\:\:\Omega\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{1}} ^{\mathrm{3}} \frac{{dt}}{{t}\:}\:\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{log}\:{t}\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\Omega\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}\:\mathrm{3} \\ $$
Answered by floor(10²Eta[1]) last updated on 18/Jun/22
![t=tg((x/2)) cos(x)=1−2sin^2 ((x/2)) 1−(2/(csc^2 ((x/2))))=1−(2/(1+cotg^2 ((x/2)))) 1−(2/(1+(1/t^2 )))=1−((2t^2 )/(t^2 +1))=((1−t^2 )/(1+t^2 )) ⇒sin(x)=((2t)/(1+t^2 )) ⇒((tgx)/(cos^2 x+2sin^2 x))=((tgx)/(1+sin^2 x))=((tgx)/(2−cos^2 x)) =((sinx)/(2cosx−cos^3 x))=(((2t)/(1+t^2 ))/(2(((1−t^2 )/(1+t^2 )))−(((1−t^2 )/(1+t^2 )))^3 )) =((2t(1+t^2 )^2 )/(2(1−t^2 )(1+t^2 )^2 −(1−t^2 )^3 ))=((2t(1+t^2 )^2 )/((1−t^2 )[1+6t^2 +t^4 ])) ⇒∫_0 ^(tg(π/8)) ((2t(1+t^2 )^2 )/((1−t^2 )(t^4 +6t^2 +1)))dx t^2 =u 2tdt=du ∫_0 ^(tg^2 (π/8)) (((1+u)^2 )/((1−u)(u^2 +6u+1)))du =(1/2)∫_0 ^(tg^2 (π/8)) (du/(1−u))−(1/2)∫_0 ^(tg^2 (π/8)) ((u−1)/(u^2 +6u+1))du =−(1/2)ln∣1−u∣_0 ^(tg^2 (π/8)) −(1/2)∫_0 ^(tg^2 (π/8)) ((u−1)/((u+3)^2 −8))du =−(1/2)ln∣1−tg^2 (π/8)∣−(1/2)∫_0 ^(tg^2 (π/8)) ((u−1)/((u+3+2(√2))(u+3−2(√2))))du =−(1/2)ln∣1−tg^2 (π/8)∣−(1/2)(((1+(√2))/2))∫_0 ^(tg^2 (π/8)) (du/(u+3+2(√2)))−(1/2)(((1−(√2))/2))∫_0 ^(tg^2 (π/8)) (du/(u+3−2(√2))) =−(1/2)ln∣1−tg^2 (π/8)∣−(1/2)(((1+(√2))/2))∫_0 ^(tg^2 (π/8)) (du/(u+3+2(√2)))−(1/2)(((1−(√2))/2))∫_0 ^(tg^2 (π/8)) (du/(u+3−2(√2))) =−(1/2)ln∣1−tg^2 (π/8)∣−((1+(√2))/4)[ln∣u+3+2(√2)∣]_0 ^(tg^2 (π/8)) −((1−(√2))/4)[ln∣u+3−2(√2)∣]_0 ^(tg^2 (π/8)) =−(1/2)ln∣1−tg^2 (π/8)∣−((1+(√2))/4)ln∣tg^2 (π/8)+3+2(√2)∣−((1−(√2))/4)ln∣tg^2 (π/8)+3−2(√2)∣+((√2)/2)ln(3+2(√2))](https://www.tinkutara.com/question/Q171571.png)
$$\mathrm{t}=\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{cos}\left(\mathrm{x}\right)=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}−\frac{\mathrm{2}}{\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cotg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}=\mathrm{1}−\frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{sin}\left(\mathrm{x}\right)=\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$ \\ $$$$\Rightarrow\frac{\mathrm{tgx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{2sin}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{tgx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{tgx}}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\frac{\mathrm{sinx}}{\mathrm{2cosx}−\mathrm{cos}^{\mathrm{3}} \mathrm{x}}=\frac{\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{2}\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)−\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2t}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{2t}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left[\mathrm{1}+\mathrm{6t}^{\mathrm{2}} +\mathrm{t}^{\mathrm{4}} \right]} \\ $$$$ \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{tg}\frac{\pi}{\mathrm{8}}} \frac{\mathrm{2t}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{t}^{\mathrm{4}} +\mathrm{6t}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$\mathrm{t}^{\mathrm{2}} =\mathrm{u} \\ $$$$\mathrm{2tdt}=\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{u}\right)\left(\mathrm{u}^{\mathrm{2}} +\mathrm{6u}+\mathrm{1}\right)}\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{du}}{\mathrm{1}−\mathrm{u}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{u}−\mathrm{1}}{\mathrm{u}^{\mathrm{2}} +\mathrm{6u}+\mathrm{1}}\mathrm{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}−\mathrm{u}\mid_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{u}−\mathrm{1}}{\left(\mathrm{u}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{8}}\mathrm{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\mid−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{u}−\mathrm{1}}{\left(\mathrm{u}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{u}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\mathrm{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\mid−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{du}}{\mathrm{u}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{du}}{\mathrm{u}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\mid−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{du}}{\mathrm{u}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\int_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \frac{\mathrm{du}}{\mathrm{u}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\mid−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{4}}\left[\mathrm{ln}\mid\mathrm{u}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\mid\right]_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} −\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{4}}\left[\mathrm{ln}\mid\mathrm{u}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\mid\right]_{\mathrm{0}} ^{\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\mid−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\mid−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\mid+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$