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Question Number 123552 by Lordose last updated on 26/Nov/20

∫_( 0) ^( (π/4)) tan^n (x)dx

$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \mathrm{tan}^{\mathrm{n}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$

Answered by Adeleke last updated on 26/Nov/20

Answered by TANMAY PANACEA last updated on 26/Nov/20

I_n =∫_0 ^(π/4) tan^(n−2) x(sec^2 x−1)dx =∫_0 ^(π/4) tan^(n−2) x×d(tanx)−I_(n−2) =∣((tan^(n−1) x)/(n−1))∣_0 ^(π/4) −I_(n−2) =(1/(n−1))−I_(n−2) I_n =(1/(n−1))−I_(n−2)

$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{n}−\mathrm{2}} {x}\left({sec}^{\mathrm{2}} {x}−\mathrm{1}\right){dx} \\ $$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{n}−\mathrm{2}} {x}×{d}\left({tanx}\right)−{I}_{{n}−\mathrm{2}} \\ $$ $$=\mid\frac{{tan}^{{n}−\mathrm{1}} {x}}{{n}−\mathrm{1}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −{I}_{{n}−\mathrm{2}} \\ $$ $$=\frac{\mathrm{1}}{{n}−\mathrm{1}}−{I}_{{n}−\mathrm{2}} \\ $$ $${I}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}−{I}_{{n}−\mathrm{2}} \\ $$

Commented bypeter frank last updated on 26/Nov/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Dwaipayan Shikari last updated on 26/Nov/20

If it was∫_0 ^(π/2) tan^n x dx then we can get a closed form ∫_0 ^(π/2) sin^n x cos^(−n) x dx =(1/2)∫_0 ^1 t^((n/2)−(1/2)) (1−t)^(((−n)/2)−(1/2)) dt sin^2 x=t =(1/2) ((Γ((n/2)+(1/2))Γ((1/2)−(n/2)))/(Γ(1)))=(π/(2sin(((nπ)/2)+(π/2)))) ∫_0 ^(π/4) tan^p x dx =∫_0 ^1 (t^p /(t^2 +1))dt =∫_0 ^1 t^p Σ_(n=0) ^∞ (−1)^n t^(2n) =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^(2n+p) dt=Σ_(n=0) ^∞ (−1)^n (1/(2n+p+1))

$${If}\:{it}\:{was}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}^{{n}} {x}\:{dx} \\ $$ $${then}\:{we}\:{can}\:{get}\:{a}\:{closed}\:{form} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {x}\:{cos}^{−{n}} {x}\:{dx} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{t}\right)^{\frac{−{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} {dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}^{\mathrm{2}} {x}={t} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{n}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}\right)}=\frac{\pi}{\mathrm{2}{sin}\left(\frac{{n}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{p}} {x}\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{p}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{p}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}+{p}} {dt}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{\mathrm{2}{n}+{p}+\mathrm{1}} \\ $$

Answered by mnjuly1970 last updated on 26/Nov/20

Ω=^(tan(x)=y) ∫_0 ^( 1) ((y^n dy)/(1+y^2 ))=∫_0 ^( 1) ((y^n −y^(n+2) )/(1−y^4 ))dy =(1/4)∫_0 ^( 1) ((t^((n−3)/4) −t^((n−1)/4) )/(1−t))dt =(1/4)(H_((n−1)/4) −H_((n−3)/4) ) =(1/4)(ψ((n/4)+(3/4))−ψ((n/4)+(1/4))) ✓

$$\Omega\overset{{tan}\left({x}\right)={y}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{y}^{{n}} {dy}}{\mathrm{1}+{y}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{y}^{{n}} −{y}^{{n}+\mathrm{2}} }{\mathrm{1}−{y}^{\mathrm{4}} }{dy} \\ $$ $$\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{t}^{\frac{{n}−\mathrm{3}}{\mathrm{4}}} \:−{t}^{\frac{{n}−\mathrm{1}}{\mathrm{4}}} }{\mathrm{1}−{t}}{dt} \\ $$ $$\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{H}_{\frac{{n}−\mathrm{1}}{\mathrm{4}}} −\mathrm{H}_{\frac{{n}−\mathrm{3}}{\mathrm{4}}} \right) \\ $$ $$\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{{n}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{{n}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right)\:\checkmark \\ $$ $$ \\ $$ $$ \\ $$ $$\: \\ $$ $$\:\: \\ $$ $$\: \\ $$

Commented byDwaipayan Shikari last updated on 26/Nov/20

∫_0 ^(π/4) tan^4 x dx=(1/4)(ψ(1+(3/4))−ψ(1+(1/4))) =(1/4)(ψ((3/4))+(4/3)−ψ((1/4))−4) = (1/4)(πcot((π/4))−(8/3))=(π/4)−(2/3)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{\mathrm{4}} {x}\:{dx}=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)+\frac{\mathrm{4}}{\mathrm{3}}−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{4}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\pi{cot}\left(\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{8}}{\mathrm{3}}\right)=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Commented byLordose last updated on 02/Dec/20

Exactly

$$\mathrm{Exactly} \\ $$

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