Question Number 206224 by mathzup last updated on 09/Apr/24
$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{5}} \right){dx} \\ $$
Answered by Frix last updated on 10/Apr/24
![∫tan^(−1) x^n dx =_(u′=1∧v=tan^(−1) x^n ) ^(by parts) =xtan^(−1) x^n −n∫(x^n /(x^(2n) +1))dx With n=5 and x∈[0, 1] we get (((√5)ln (1+(√5)))/2)−(((1+(√5))ln 2)/2)+((5−(√(10(5−(√5)))))/(20))π≈.151019250](https://www.tinkutara.com/question/Q206239.png)
$$\int\mathrm{tan}^{−\mathrm{1}} \:{x}^{{n}} \:{dx}\:\underset{{u}'=\mathrm{1}\wedge{v}=\mathrm{tan}^{−\mathrm{1}} \:{x}^{{n}} } {\overset{\mathrm{by}\:\mathrm{parts}} {=}}\: \\ $$$$={x}\mathrm{tan}^{−\mathrm{1}} \:{x}^{{n}} \:−{n}\int\frac{{x}^{{n}} }{{x}^{\mathrm{2}{n}} +\mathrm{1}}{dx} \\ $$$$\mathrm{With}\:{n}=\mathrm{5}\:\mathrm{and}\:{x}\in\left[\mathrm{0},\:\mathrm{1}\right]\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\sqrt{\mathrm{5}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)}{\mathrm{2}}−\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}+\frac{\mathrm{5}−\sqrt{\mathrm{10}\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)}}{\mathrm{20}}\pi\approx.\mathrm{151019250} \\ $$
Answered by mathzup last updated on 10/Apr/24
$${thank}\:{you}\:{sir} \\ $$
Answered by mathzup last updated on 10/Apr/24
![by parts we get I=[x arctan(x^5 )]_0 ^1 −∫_0 ^1 x.((5x^4 )/(1+x^(10) ))dx =(π/4)−5∫_0 ^1 (x^5 /(1+x^(10) ))dx but ∫_0 ^1 (x^5 /(1+x^(10) ))dx=∫_0 ^1 x^5 Σ(−1)^n x^(10n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(10n+5) dx =Σ_(n=0) ^∞ (−1)^n ×(1/(10n+6)) =Σ_(n=o) ^∞ (1/(10(2n)+6))−Σ_(n=0) ^∞ (1/(10(2n+1)+6)) =Σ_(n=0) ^∞ ((1/(20n+6))−(1/(20n+16))) =10Σ_(n=0) ^∞ (1/((20n+6)(20n+16))) =((10)/(20^2 ))Σ_(n=0) ^∞ (1/((n+(6/(20)))(n+((16)/(20))))) =(1/(40))×((Ψ(((16)/(20)))−Ψ((6/(20))))/(((16)/(20))−(6/(20))))=(1/(20)){Ψ((4/5))−Ψ((3/(10)))}](https://www.tinkutara.com/question/Q206266.png)
$${by}\:{parts}\:{we}\:{get} \\ $$$${I}=\left[{x}\:{arctan}\left({x}^{\mathrm{5}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {x}.\frac{\mathrm{5}{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{10}} }{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}−\mathrm{5}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{5}} }{\mathrm{1}+{x}^{\mathrm{10}} }{dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{5}} }{\mathrm{1}+{x}^{\mathrm{10}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{5}} \Sigma\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{10}{n}} \:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{10}{n}+\mathrm{5}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} ×\frac{\mathrm{1}}{\mathrm{10}{n}+\mathrm{6}} \\ $$$$=\sum_{{n}={o}} ^{\infty} \frac{\mathrm{1}}{\mathrm{10}\left(\mathrm{2}{n}\right)+\mathrm{6}}−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{10}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{6}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{20}{n}+\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{20}{n}+\mathrm{16}}\right) \\ $$$$=\mathrm{10}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{20}{n}+\mathrm{6}\right)\left(\mathrm{20}{n}+\mathrm{16}\right)} \\ $$$$=\frac{\mathrm{10}}{\mathrm{20}^{\mathrm{2}} }\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{6}}{\mathrm{20}}\right)\left({n}+\frac{\mathrm{16}}{\mathrm{20}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{40}}×\frac{\Psi\left(\frac{\mathrm{16}}{\mathrm{20}}\right)−\Psi\left(\frac{\mathrm{6}}{\mathrm{20}}\right)}{\frac{\mathrm{16}}{\mathrm{20}}−\frac{\mathrm{6}}{\mathrm{20}}}=\frac{\mathrm{1}}{\mathrm{20}}\left\{\Psi\left(\frac{\mathrm{4}}{\mathrm{5}}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{10}}\right)\right\} \\ $$