Question Number 156743 by cortano last updated on 15/Oct/21
Answered by MJS_new last updated on 15/Oct/21
![0≤x≤(1/2) ⇒ ∫(dx/( (((1+x)^2 (1−x)^6 ))^(1/4) ))= =∫(dx/( (1−x)(√(1−x^2 ))))= [t=((1+(√(1−x^2 )))/x) → dx=−((x^2 (√(1−x^2 )))/(1+(√(1−x^2 ))))dt] =−2∫(dt/((t−1)^2 ))=(2/(t−1))=... =((√(1−x^2 ))/(1−x))−1+C ⇒ (−1+(√3))∫_0 ^(1/2) (dx/( (((1+x)^2 (1−x)^6 ))^(1/4) ))=4−2(√3)](https://www.tinkutara.com/question/Q156760.png)
$$\mathrm{0}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\int\frac{{dx}}{\:\sqrt[{\mathrm{4}}]{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{6}} }}= \\ $$$$=\int\frac{{dx}}{\:\left(\mathrm{1}−{x}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\rightarrow\:{dx}=−\frac{{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dt}\right] \\ $$$$=−\mathrm{2}\int\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2}}{{t}−\mathrm{1}}=… \\ $$$$=\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}−{x}}−\mathrm{1}+{C} \\ $$$$\Rightarrow \\ $$$$\left(−\mathrm{1}+\sqrt{\mathrm{3}}\right)\underset{\mathrm{0}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\frac{{dx}}{\:\sqrt[{\mathrm{4}}]{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{6}} }}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$
Answered by FongXD last updated on 15/Oct/21
![=∫_0 ^(1/2) ((((√3)−1)/( (((1+x)^2 (1−x)^2 (1−x)^4 ))^(1/4) )))dx =∫_0 ^(1/2) [(((√3)−1)/((1−x)(((1−x^2 )^2 ))^(1/4) ))]dx let x=sinθ, ⇒ dx=cosθdθ =∫_0 ^(π/6) (((√3)−1)/((1−sinθ)(((1−sin^2 θ)^2 ))^(1/4) ))×cosθdθ =∫_0 ^(π/6) ((((√3)−1)dθ)/(1−sinθ))=∫_0 ^(π/6) ((((√3)−1)dθ)/((cos(θ/2)−sin(θ/2))^2 )) =(((√3)−1)/2)∫_0 ^(π/6) (dθ/(cos^2 ((θ/2)+(π/4))))=(((√3)−1)/2)[2tan((θ/2)+(π/4))]_0 ^(π/6) =((√3)−1)(tan(π/3)−tan(π/4))=((√3)−1)^2 =4−2(√3) so. determinant (((∫_0 ^(1/2) ((((√3)−1)/( (((1+x)^2 (1−x)^4 ))^(1/4) )))dx=4−2(√3))))](https://www.tinkutara.com/question/Q156753.png)
$$=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} }}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)\sqrt[{\mathrm{4}}]{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }}\right]\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{sin}\theta,\:\Rightarrow\:\mathrm{dx}=\mathrm{cos}\theta\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\left(\mathrm{1}−\mathrm{sin}\theta\right)\sqrt[{\mathrm{4}}]{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }}×\mathrm{cos}\theta\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\mathrm{d}\theta}{\mathrm{1}−\mathrm{sin}\theta}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\mathrm{d}\theta}{\left(\mathrm{cos}\frac{\theta}{\mathrm{2}}−\mathrm{sin}\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \frac{\mathrm{d}\theta}{\mathrm{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\left[\mathrm{2tan}\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \\ $$$$=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\mathrm{tan}\frac{\pi}{\mathrm{3}}−\mathrm{tan}\frac{\pi}{\mathrm{4}}\right)=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\mathrm{so}.\:\begin{array}{|c|}{\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} }}\right)\mathrm{dx}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}\\\hline\end{array} \\ $$
Commented by cortano last updated on 15/Oct/21
$$\mathrm{yes} \\ $$