Question Number 109017 by mathdave last updated on 20/Aug/20
Answered by mathmax by abdo last updated on 20/Aug/20
![I =∫_(arctan(1)) ^(arctan((√2))) (√(8/(1−sin(4x))))dx ⇒I =∫_(arctan(1)) ^(arctan((√2))) (√(8/(1−cos((π/2)−4x))))dx =∫_(arctan(1)) ^(arctan((√2))) (√(8/(2sin^2 ((π/4)−2x))))dx =∫_(arctan(1)) ^(arctan((√2))) (2/(sin((π/4)−2x)))dx =_((π/4)−2x =t) 2∫_((π/4)−2arctan(1)) ^((π/4)−2arctan((√2))) ((−dt)/(2 sin(t))) =∫_((π/4)−2arctan((√2))) ^(−(π/4)) (dt/(sint)) =_(tan((t/2))=u) ∫_(tan((π/8)−arctan(√2))) ^(−((√2)−1)) ((2du)/((1+u^2 )((2u)/(1+u^2 )))) =∫_(tan((π/8)−arctan(√2))) ^(1−(√2)) (du/u) =[ln∣u∣]_(tan((π/8)−arctan((√2)))) ^(1−(√2)) =ln((√2)−1)−ln∣tan((π/8)−arctan((√2)))∣](https://www.tinkutara.com/question/Q109077.png)
$$\mathrm{I}\:=\int_{\mathrm{arctan}\left(\mathrm{1}\right)} ^{\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)} \sqrt{\frac{\mathrm{8}}{\mathrm{1}−\mathrm{sin}\left(\mathrm{4x}\right)}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{arctan}\left(\mathrm{1}\right)} ^{\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)} \sqrt{\frac{\mathrm{8}}{\mathrm{1}−\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{4x}\right)}}\mathrm{dx} \\ $$$$=\int_{\mathrm{arctan}\left(\mathrm{1}\right)} ^{\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)} \sqrt{\frac{\mathrm{8}}{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right)}}\mathrm{dx}\:=\int_{\mathrm{arctan}\left(\mathrm{1}\right)} ^{\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)} \frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right)}\mathrm{dx} \\ $$$$=_{\frac{\pi}{\mathrm{4}}−\mathrm{2x}\:=\mathrm{t}} \:\:\:\:\mathrm{2}\int_{\frac{\pi}{\mathrm{4}}−\mathrm{2arctan}\left(\mathrm{1}\right)} ^{\frac{\pi}{\mathrm{4}}−\mathrm{2arctan}\left(\sqrt{\mathrm{2}}\right)} \:\:\:\:\:\frac{−\mathrm{dt}}{\mathrm{2}\:\mathrm{sin}\left(\mathrm{t}\right)} \\ $$$$=\int_{\frac{\pi}{\mathrm{4}}−\mathrm{2arctan}\left(\sqrt{\mathrm{2}}\right)} ^{−\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{dt}}{\mathrm{sint}}\:=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{u}} \:\:\:\:\int_{\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}−\mathrm{arctan}\sqrt{\mathrm{2}}\right)} ^{−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} \:\:\:\:\frac{\mathrm{2du}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\frac{\mathrm{2u}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }} \\ $$$$=\int_{\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}−\mathrm{arctan}\sqrt{\mathrm{2}}\right)} ^{\mathrm{1}−\sqrt{\mathrm{2}}} \:\frac{\mathrm{du}}{\mathrm{u}}\:\:=\left[\mathrm{ln}\mid\mathrm{u}\mid\right]_{\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}−\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)\right)} ^{\mathrm{1}−\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−\mathrm{ln}\mid\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}−\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)\right)\mid \\ $$