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Question Number 81801 by Power last updated on 15/Feb/20

Commented by mathmax by abdo last updated on 16/Feb/20

changement x^n =cos^2 t give x =(cost)^(2/n) ⇒dx =−(2/n)sint (cost)^((2/n)−1) ⇒∫_0 ^1 (dx/((^n (√(1−x^n ))))) =(2/n)∫_0 ^(π/2) ((sint)/((^n (√(1−cos^2 t)))))(cost)^((2/n)−1) dt =(2/n) ∫_0 ^(π/2) ((sint)/(sint^(2/n) )) (cost)^((2/n)−1) dt =(2/n) ∫_0 ^(π/2) (sint)^(1−(2/n)) (cost)^((2/n)−1) dt we have the formula 2 ∫_0 ^(π/2) (cost)^(2p−1) (sint)^(2q−1) =B(p,q) =((Γ(p)Γ(q))/(Γ(p+q))) ⇒∫_0 ^1 (dx/((^n (√(1−x^n ))))) =(2/n)∫_0 ^(π/2) (sint)^(2(1−(1/n))−1) (cost)^((2/n)−1) dt =(1/n) B(1−(1/n),(1/n)) =(1/n)×((Γ(1−(1/n))Γ((1/n)))/(Γ(1−(1/n) +(1/n))))=(1/n)×(π/(sin((π/n))))=(π/(nsin((π/n))))

$${changement}\:{x}^{{n}} \:={cos}^{\mathrm{2}} {t}\:{give}\:{x}\:=\left({cost}\right)^{\frac{\mathrm{2}}{{n}}} \:\Rightarrow{dx}\:=−\frac{\mathrm{2}}{{n}}{sint}\:\left({cost}\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left(^{{n}} \sqrt{\mathrm{1}−{x}^{{n}} }\right)}\:=\frac{\mathrm{2}}{{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sint}}{\left(^{{n}} \sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {t}}\right)}\left({cost}\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} \:{dt} \\ $$$$=\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sint}}{{sint}^{\frac{\mathrm{2}}{{n}}} }\:\left({cost}\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} \:{dt}\:=\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({sint}\right)^{\mathrm{1}−\frac{\mathrm{2}}{{n}}} \:\left({cost}\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} \:{dt} \\ $$$${we}\:{have}\:{the}\:{formula}\:\:\:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cost}\right)^{\mathrm{2}{p}−\mathrm{1}} \:\left({sint}\right)^{\mathrm{2}{q}−\mathrm{1}} \:={B}\left({p},{q}\right) \\ $$$$=\frac{\Gamma\left({p}\right)\Gamma\left({q}\right)}{\Gamma\left({p}+{q}\right)}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left(^{{n}} \sqrt{\left.\mathrm{1}−{x}^{{n}} \right)}\right.}\:=\frac{\mathrm{2}}{{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({sint}\right)^{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)−\mathrm{1}} \left({cost}\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\:{B}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}},\frac{\mathrm{1}}{{n}}\right)\:=\frac{\mathrm{1}}{{n}}×\frac{\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\frac{\mathrm{1}}{{n}}\right)}{\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{{n}}\right)}=\frac{\mathrm{1}}{{n}}×\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)}=\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)} \\ $$

Answered by TANMAY PANACEA last updated on 15/Feb/20

x^n =sin^2 θ→x=(sinθ)^(2/n) dx=(2/n)(sinθ)^((2/n)−1) cosθ dθ ∫_0 ^(π/2) (((2/n)(sinθ)^((2/n)−1) cosθdθ)/((cosθ)^(2/n) )) (1/n)×2∫_0 ^(π/2) (sinθ)^((2/n)−1) .(cosθ)^(1−(2/n)) dθ formula 2∫_0 ^(π/2) (sinθ)^(2p−1) (cosθ)^(2q−1) dθ=((⌈(p)⌈(q))/(⌈(p+q))) 2p−1=(2/n)−1→p=(1/n) 2q−1=1−(2/(n ))→q=1−(1/n) p+q=1 so =((⌈((1/n))⌈(1−(1/n)))/(⌈(1)))=(π/(sin((π/n)))) answer is =(1/n)×(π/(sin((π/n)))) formula ⌈(p)⌈(1−p)=(π/(sin(pπ))) pls check mistake if any

$${x}^{{n}} ={sin}^{\mathrm{2}} \theta\rightarrow{x}=\left({sin}\theta\right)^{\frac{\mathrm{2}}{{n}}} \\ $$$${dx}=\frac{\mathrm{2}}{{n}}\left({sin}\theta\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} {cos}\theta\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\mathrm{2}}{{n}}\left({sin}\theta\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} {cos}\theta{d}\theta}{\left({cos}\theta\right)^{\frac{\mathrm{2}}{{n}}} } \\ $$$$\frac{\mathrm{1}}{{n}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\frac{\mathrm{2}}{{n}}−\mathrm{1}} .\left({cos}\theta\right)^{\mathrm{1}−\frac{\mathrm{2}}{{n}}} {d}\theta \\ $$$${formula}\: \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}{p}−\mathrm{1}} \left({cos}\theta\right)^{\mathrm{2}{q}−\mathrm{1}} {d}\theta=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\ $$$$\mathrm{2}{p}−\mathrm{1}=\frac{\mathrm{2}}{{n}}−\mathrm{1}\rightarrow{p}=\frac{\mathrm{1}}{{n}} \\ $$$$\mathrm{2}{q}−\mathrm{1}=\mathrm{1}−\frac{\mathrm{2}}{{n}\:}\rightarrow{q}=\mathrm{1}−\frac{\mathrm{1}}{{n}} \\ $$$${p}+{q}=\mathrm{1} \\ $$$${so}\:=\frac{\lceil\left(\frac{\mathrm{1}}{{n}}\right)\lceil\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)}{\lceil\left(\mathrm{1}\right)}=\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)} \\ $$$${answer}\:{is}\:=\frac{\mathrm{1}}{{n}}×\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)} \\ $$$$ \\ $$$${formula}\:\lceil\left({p}\right)\lceil\left(\mathrm{1}−{p}\right)=\frac{\pi}{{sin}\left({p}\pi\right)} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}}\:\boldsymbol{{mistake}}\:\boldsymbol{{if}}\:\boldsymbol{{any}} \\ $$

Commented by Power last updated on 15/Feb/20

⌈ − ???

$$\lceil\:−\:??? \\ $$

Commented by Tony Lin last updated on 15/Feb/20

Γ(gamma function)

$$\Gamma\left({gamma}\:{function}\right) \\ $$

Commented by TANMAY PANACEA last updated on 15/Feb/20

yes sir thank you...i forgot to mention ⌈ represent gamma function

$${yes}\:{sir}\:{thank}\:{you}...{i}\:{forgot}\:{to}\:{mention}\:\lceil\:{represent} \\ $$$${gamma}\:{function} \\ $$

Commented by Power last updated on 15/Feb/20

thanks

$$\mathrm{thanks} \\ $$

Commented by mathmax by abdo last updated on 16/Feb/20

you answer is correct sir tanmay

$${you}\:{answer}\:{is}\:{correct}\:{sir}\:{tanmay} \\ $$

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