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Question Number 36753 by prof Abdo imad last updated on 05/Jun/18
find I_n = ∫_0 ^1 x^n arctan(x)dx .
$${find}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}^{{n}} \:{arctan}\left({x}\right){dx}\:. \\ $$
Commented by prof Abdo imad last updated on 05/Jun/18
we have arctan^′ (x)=(1/(1+x^2 )) =Σ_(n=0) ^∞ (−1)^n x^(2n) ⇒ arctan(x)=Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) +c (c=0)⇒ arctan(x)=Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) ⇒ I_n = ∫_0 ^1 (Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) )x^n dx = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) ∫_0 ^1 x^(3n+1) dx =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(3n+2))) let w(x) = Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(3n+2)))x^(3n+2) w^′ (x) = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(3n) = Σ_(n=0) ^∞ (((−x^3 )^n )/(2n+1)) =ϕ(−x^3 ) with ϕ(t) = Σ_(n=0) ^∞ (t^n /(2n+1)) =(1/( (√t)))Σ_(n=0) ^∞ ((((√t))^(2n+1) )/(2n+1)) but (Σ_(n=0) ^∞ (t^(2n+1) /(2n+1)))^′ =Σ_(n=0) ^∞ t^(2n) = (1/(1−t^2 )) ⇒ Σ_(n=0) ^∞ (t^(2n+1) /(2n+1)) = ∫ (dt/(1−t^2 )) =(1/2)∫ ( (1/(1−t)) +(1/(1+t)))dx =(1/2)ln∣((1+t)/(1−t))∣ +c (c=0) ⇒ ϕ(t)= (1/( (√t))) (1/2)ln∣((1+(√t))/(1−(√t)))∣ if t<0 ϕ(t) = Σ_(n=0) ^∞ (((−1)^n (−t)^n )/(2n+1)) = (1/( (√(−t)))) Σ_(n=0) ^∞ (−1)^n ((((√(−t)))^(2n+1) )/(2n+1)) ...be continued...
$${we}\:{have}\:{arctan}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} \:\Rightarrow \\ $$$${arctan}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \:+{c}\:\left({c}=\mathrm{0}\right)\Rightarrow \\ $$$${arctan}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \:\Rightarrow \\ $$$${I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \right){x}^{{n}} {dx} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{3}{n}+\mathrm{1}} \:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)} \\ $$$${let}\:{w}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}{x}^{\mathrm{3}{n}+\mathrm{2}} \\ $$$${w}^{'} \left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{3}{n}} \:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−{x}^{\mathrm{3}} \right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\varphi\left(−{x}^{\mathrm{3}} \right)\:{with}\:\varphi\left({t}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{t}}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(\sqrt{{t}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:\:\:{but} \\ $$$$\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\right)^{'} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{t}^{\mathrm{2}{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:=\:\int\:\:\:\:\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\:\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{c}\:\:\left({c}=\mathrm{0}\right)\:\Rightarrow\:\varphi\left({t}\right)=\:\frac{\mathrm{1}}{\:\sqrt{{t}}}\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+\sqrt{{t}}}{\mathrm{1}−\sqrt{{t}}}\mid \\ $$$${if}\:{t}<\mathrm{0}\:\:\varphi\left({t}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \left(−{t}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{−{t}}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:\:\frac{\left(\sqrt{−{t}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:\:…{be}\:{continued}… \\ $$
Commented by prof Abdo imad last updated on 06/Jun/18
but Σ_(n=0) ^∞ (−1)^n (u^(2n+1) /(2n+1)))^′ = Σ_(n=0) ^∞ (−1)^n u^(2n) = (1/(1+u^2 )) ⇒ Σ_(n=0) ^∞ (−1)^n (u^(2n+1) /(2n+1)) =arctanu +c(c=0) ⇒ ϕ(t) = (1/( (√(−t)))) arctan((√(−t))) w^′ (x) =ϕ(−x^3 ) = (1/( (√x^3 )))?arctan((√x^3 )) ⇒ w(x)= ∫ (1/(x(√x))) arctan(x(√x))dx +c chang. x(√x_ )=t give x=t^(2/3) w(x) = ∫ (1/t) arctan(t) (2/3)t^((2/3)−1) dt = (2/3) ∫ t^((2/3)−2) arctan(t)dt =(2/3) ∫ t^(−(4/3)) arctant dt by parts let find ∫ t^(−λ) arctant dt = (1/(1−λ)) t^(1−λ) arctant −∫ (1/(1−λ)) (t^(1−λ) /(1+t^2 )) dt ....
$$\left.{but}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\right)^{'} \:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {u}^{\mathrm{2}{n}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\Rightarrow\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:={arctanu}\:+{c}\left({c}=\mathrm{0}\right) \\ $$$$\Rightarrow\:\varphi\left({t}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{−{t}}}\:{arctan}\left(\sqrt{−{t}}\right) \\ $$$${w}^{'} \left({x}\right)\:=\varphi\left(−{x}^{\mathrm{3}} \right)\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{3}} }}?{arctan}\left(\sqrt{{x}^{\mathrm{3}} }\right)\:\Rightarrow \\ $$$${w}\left({x}\right)=\:\int\:\:\:\frac{\mathrm{1}}{{x}\sqrt{{x}}}\:{arctan}\left({x}\sqrt{{x}}\right){dx}\:+{c} \\ $$$${chang}.\:{x}\sqrt{{x}_{} }={t}\:{give}\:\:{x}={t}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$${w}\left({x}\right)\:=\:\int\:\:\frac{\mathrm{1}}{{t}}\:{arctan}\left({t}\right)\:\frac{\mathrm{2}}{\mathrm{3}}{t}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \:{dt} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:\int\:\:{t}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{2}} \:{arctan}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\:\int\:\:\:\:{t}^{−\frac{\mathrm{4}}{\mathrm{3}}} \:{arctant}\:{dt}\:\:{by}\:{parts}\:{let}\:{find} \\ $$$$\int\:{t}^{−\lambda} \:{arctant}\:{dt}\:=\:\frac{\mathrm{1}}{\mathrm{1}−\lambda}\:{t}^{\mathrm{1}−\lambda} \:{arctant} \\ $$$$−\int\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−\lambda}\:\:\:\frac{{t}^{\mathrm{1}−\lambda} }{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:…. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18
tan^(−1) x=x_ −(x^3 /3)+(x^5 /5)−(x^7 /7)+... I_n =∫_0 ^1 x^n (x−(x^3 /3)+(x^5 /5)−(x^7 /7)+...)dx =∫_0 ^1 (x^(n+1) /1)−(x^(n+3) /3)+(x^(n+5) /5)−(x^(7+n) /7)+...)dx ∣(x^(n+2) /(1×(n+2)))−(x^(n+4) /((n+4)×3))+(x^(n+6) /((n+6)×5))−(x^(n+8) /((n+8)×7))..∣_0 ^1 (1/(1×(n+2)))−(1/((n+4)×3))+(1/((n+6)×5))−(1/((n+8)×7...)) =
$$ \\ $$$${tan}^{−\mathrm{1}} {x}={x}_{} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+… \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+…\right){dx} \\ $$$$\left.=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{\mathrm{1}}−\frac{{x}^{{n}+\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{{n}+\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}+{n}} }{\mathrm{7}}+…\right){dx} \\ $$$$\mid\frac{{x}^{{n}+\mathrm{2}} }{\mathrm{1}×\left({n}+\mathrm{2}\right)}−\frac{{x}^{{n}+\mathrm{4}} }{\left({n}+\mathrm{4}\right)×\mathrm{3}}+\frac{{x}^{{n}+\mathrm{6}} }{\left({n}+\mathrm{6}\right)×\mathrm{5}}−\frac{{x}^{{n}+\mathrm{8}} }{\left({n}+\mathrm{8}\right)×\mathrm{7}}..\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}×\left({n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\left({n}+\mathrm{4}\right)×\mathrm{3}}+\frac{\mathrm{1}}{\left({n}+\mathrm{6}\right)×\mathrm{5}}−\frac{\mathrm{1}}{\left({n}+\mathrm{8}\right)×\mathrm{7}…} \\ $$$$= \\ $$

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