Question Number 108208 by bemath last updated on 15/Aug/20
$$\:\:\:\:\:\:\:\frac{\mathcal{B}{e}\mathcal{M}{ath}}{\boxminus} \\ $$$$\left(\mathrm{1}\right)\int\:\frac{{x}^{\mathrm{5}} −{x}}{{x}^{\mathrm{8}} +\mathrm{1}}\:{dx}\: \\ $$$$\left(\mathrm{2}\right)\:\underset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{3}} }\:{dx}\: \\ $$
Answered by john santu last updated on 15/Aug/20
$$\:\:\:\:\frac{\bigstar\mathcal{JS}\spadesuit}{\Bumpeq} \\ $$$${J}=\int\:\frac{{x}\left({x}^{\mathrm{4}} −\mathrm{1}\right)}{{x}^{\mathrm{8}} +\mathrm{1}}\:{dx}\: \\ $$$${let}\:{z}\:=\:{x}^{\mathrm{2}} \:\Rightarrow{J}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{4}} +\mathrm{1}}\:{dz}\: \\ $$$${decomposition}\: \\ $$$${z}^{\mathrm{4}} +\mathrm{1}\:=\:\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{z}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left({z}^{\mathrm{2}} +{z}\sqrt{\mathrm{2}}+\mathrm{1}\right)\left({z}^{\mathrm{2}} −{z}\sqrt{\mathrm{2}}\:+\mathrm{1}\right) \\ $$$$\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{4}} +\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{z}\sqrt{\mathrm{2}}−\mathrm{1}}{{z}^{\mathrm{2}} −{z}\sqrt{\mathrm{2}}+\mathrm{1}}−\frac{{z}\sqrt{\mathrm{2}}\:+\mathrm{1}}{{z}^{\mathrm{2}} +{z}\sqrt{\mathrm{2}}+\mathrm{1}}\right) \\ $$$${J}=\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\frac{{z}\sqrt{\mathrm{2}}−\mathrm{1}}{{z}^{\mathrm{2}} −{z}\sqrt{\mathrm{2}}+\mathrm{1}}−\frac{{z}\sqrt{\mathrm{2}}\:+\mathrm{1}}{{z}^{\mathrm{2}} +{z}\sqrt{\mathrm{2}}+\mathrm{1}}\right){dz} \\ $$$${J}=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\left(\frac{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \sqrt{\mathrm{2}}\:+\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \sqrt{\mathrm{2}}+\mathrm{1}}\right)\:+\:{c}\: \\ $$
Commented by bobhans last updated on 15/Aug/20
$$\frac{\mathcal{C}\mathrm{ooll}}{\mathrm{great}}\: \\ $$
Answered by bobhans last updated on 15/Aug/20
![(2) ((bobhans)/(°•≡•°)) I = ∫_(1/(√2)) ^1 ((sin^(−1) (x))/x^3 ) dx by part { ((u=sin^(−1) (x)→du=(dx/( (√(1−x^2 )))))),((v = −(1/(2x^2 )))) :} I=[−((sin^(−1) (x))/(2x^2 ))]_(1/( (√2))) ^1 +∫_(1/(√2)) ^1 (dx/(x^2 (√(1−x^2 )))) I=0 +(1/2)∫_(π/4) ^(π/2) ((cos h dh )/(sin^2 h (√(1−sin^2 h)))) I=(1/2)∫_(π/4) ^(π/2) csc^2 h dh = −(1/2) [cot h ]_(π/4) ^(π/2) I=(1/2)](https://www.tinkutara.com/question/Q108214.png)
$$\left(\mathrm{2}\right)\:\:\:\frac{\mathrm{bobhans}}{°\bullet\equiv\bullet°} \\ $$$$\:\:\mathrm{I}\:=\:\underset{\mathrm{1}/\sqrt{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{3}} }\:\mathrm{dx}\: \\ $$$$\mathrm{by}\:\mathrm{part}\:\begin{cases}{\mathrm{u}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\rightarrow\mathrm{du}=\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}}\\{\mathrm{v}\:=\:−\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }}\end{cases} \\ $$$$\mathrm{I}=\left[−\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{2x}^{\mathrm{2}} }\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\mathrm{1}} +\underset{\mathrm{1}/\sqrt{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\: \\ $$$$\mathrm{I}=\mathrm{0}\:+\frac{\mathrm{1}}{\mathrm{2}}\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{cos}\:\mathrm{h}\:\mathrm{dh}\:}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{h}\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{h}}} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{csc}^{\mathrm{2}} \:\mathrm{h}\:\mathrm{dh}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\left[\mathrm{cot}\:\mathrm{h}\:\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 15/Aug/20
![2) I =∫_(1/( (√2))) ^1 ((arcsinx)/x^3 )dx by parts u^′ =x^(−3 ) and v =arcsinx ⇒I =[−(1/2)x^(−2) arcsinx]_(1/( (√2))) ^1 +(1/2)∫_(1/( (√2))) ^1 x^(−2) (dx/( (√(1−x^2 )))) =(π/4)−(π/4) +(1/2)∫_(1/( (√2))) ^1 (dx/(x^2 (√(1−x^2 )))) =(1/2)∫_(1/( (√2))) ^1 (dx/(x^2 (√(1−x^2 )))) =_(x=sint) (1/2)∫_(π/4) ^(π/2) ((cost dt)/(sin^2 t cost)) =(1/2)∫_(π/4) ^(π/2) ((2dt)/(1−cos(2t))) =_(2t=u) ∫_(π/2) ^π (du/(2{1−cosu})) =_(tan((u/2))=z) (1/2) ∫_1 ^(+∞) ((2dz)/((1+z^2 )(1−((1−z^2 )/(1+z^2 ))))) =∫_1 ^(+∞) (dz/(1+z^2 −1+z^2 )) =(1/2)∫_1 ^(+∞) (dz/z^2 ) =(1/2)[−(1/z)]_1 ^(+∞) =(1/2) ⇒ I =(1/2)](https://www.tinkutara.com/question/Q108233.png)
$$\left.\mathrm{2}\right)\:\mathrm{I}\:=\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\frac{\mathrm{arcsinx}}{\mathrm{x}^{\mathrm{3}} }\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{u}^{'} \:=\mathrm{x}^{−\mathrm{3}\:} \:\mathrm{and}\:\mathrm{v}\:=\mathrm{arcsinx} \\ $$$$\Rightarrow\mathrm{I}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{−\mathrm{2}} \:\mathrm{arcsinx}\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\mathrm{x}^{−\mathrm{2}} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$=\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$=_{\mathrm{x}=\mathrm{sint}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{cost}\:\mathrm{dt}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\mathrm{cost}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{2dt}}{\mathrm{1}−\mathrm{cos}\left(\mathrm{2t}\right)} \\ $$$$=_{\mathrm{2t}=\mathrm{u}} \:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\:\:\frac{\mathrm{du}}{\mathrm{2}\left\{\mathrm{1}−\mathrm{cosu}\right\}}\:=_{\mathrm{tan}\left(\frac{\mathrm{u}}{\mathrm{2}}\right)=\mathrm{z}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{\mathrm{2dz}}{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{dz}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} −\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{dz}}{\mathrm{z}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\left[−\frac{\mathrm{1}}{\mathrm{z}}\right]_{\mathrm{1}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 15/Aug/20
![1) complex method let decompose F(x) =((x^5 −x)/(x^8 +1)) x^8 +1=0 ⇒x^8 =e^(i(2k+1)π ) ⇒z_k =e^((i(2k+1)π)/8) and k∈[[0,7]] ⇒F(x) =((x^5 −x)/(Π_(k=0) ^7 (x−z_k ))) =Σ_(k=0) ^7 (a_k /(x−z_k )) a_k =((z_k ^5 −z_k )/(8z_k ^7 )) =((z_k ^6 −z_k ^2 )/(−8)) ⇒∫ F(x)dx =−(1/8)∫ Σ_(k=0) ^7 ((z_k ^6 −z_k ^2 )/(x−z_k ))dx =−(1/8)Σ_(k=0) ^7 (z_k ^6 −z_k ^2 )ln(x−z_k ) +C](https://www.tinkutara.com/question/Q108234.png)
$$\left.\mathrm{1}\right)\:\mathrm{complex}\:\mathrm{method}\:\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}^{\mathrm{5}} −\mathrm{x}}{\mathrm{x}^{\mathrm{8}} \:+\mathrm{1}} \\ $$$$\mathrm{x}^{\mathrm{8}} +\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{x}^{\mathrm{8}} \:=\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi\:} \:\Rightarrow\mathrm{z}_{\mathrm{k}} =\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{8}}} \:\:\mathrm{and}\:\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{7}\right]\right] \\ $$$$\Rightarrow\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}^{\mathrm{5}} −\mathrm{x}}{\prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \left(\mathrm{x}−\mathrm{z}_{\mathrm{k}} \right)}\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{x}−\mathrm{z}_{\mathrm{k}} } \\ $$$$\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{5}} −\mathrm{z}_{\mathrm{k}} }{\mathrm{8z}_{\mathrm{k}} ^{\mathrm{7}} }\:=\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{6}} −\mathrm{z}_{\mathrm{k}} ^{\mathrm{2}} }{−\mathrm{8}}\:\Rightarrow\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{8}}\int\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{6}} −\mathrm{z}_{\mathrm{k}} ^{\mathrm{2}} }{\mathrm{x}−\mathrm{z}_{\mathrm{k}} }\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\left(\mathrm{z}_{\mathrm{k}} ^{\mathrm{6}} −\mathrm{z}_{\mathrm{k}} ^{\mathrm{2}} \right)\mathrm{ln}\left(\mathrm{x}−\mathrm{z}_{\mathrm{k}} \right)\:+\mathrm{C} \\ $$