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Question Number 91735 by frc2crc last updated on 02/May/20

∫_0 ^(π/2) ((tan^m α))^(1/k) dα for m>1

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \sqrt[{{k}}]{\mathrm{tan}^{{m}} \:\alpha}\:{d}\alpha\:{for}\:{m}>\mathrm{1} \\ $$

Commented bymathmax by abdo last updated on 03/May/20

I =∫_0 ^(π/2) (tan^m α)^(1/k) dα chsngement tanα =x give I =∫_0 ^∞ (x^m )^(1/k) (dx/(1+x^2 )) =∫_0 ^∞ (x^(m/k) /(1+x^2 ))dx changement x=u^(1/2) give I =∫_0 ^∞ (u^(m/(2k)) /(1+u)) (1/2)u^((1/2)−1) du =(1/2)∫_0 ^∞ (u^((m/(2k))+(1/2)−1) /(1+u)) du we have proved that ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa))) if o<a<1 ⇒ I =(1/2)×(π/(sin(π((m/(2k))+(1/2))))) =(π/(2sin(((mπ)/(2k))+(π/2)))) ⇒ I =(π/(2cos(((mπ)/(2k))))) (m/(2k))+(1/2)−1 =(m/(2k))−(1/2) =(1/2)((m/k)−1)<0 ⇒m<k so the condition is 1<m<k

$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({tan}^{{m}} \alpha\right)^{\frac{\mathrm{1}}{{k}}} \:{d}\alpha\:\:{chsngement}\:{tan}\alpha\:={x}\:{give} \\ $$ $${I}\:=\int_{\mathrm{0}} ^{\infty} \left({x}^{{m}} \right)^{\frac{\mathrm{1}}{{k}}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{{m}}{{k}}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{changement}\:\:{x}={u}^{\frac{\mathrm{1}}{\mathrm{2}}} \:{give} \\ $$ $${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\frac{{m}}{\mathrm{2}{k}}} }{\mathrm{1}+{u}}\:\frac{\mathrm{1}}{\mathrm{2}}{u}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\frac{{m}}{\mathrm{2}{k}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{u}}\:{du} \\ $$ $${we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{if}\:{o}<{a}<\mathrm{1}\:\Rightarrow\: \\ $$ $${I}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{{sin}\left(\pi\left(\frac{{m}}{\mathrm{2}{k}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}\:=\frac{\pi}{\mathrm{2}{sin}\left(\frac{{m}\pi}{\mathrm{2}{k}}+\frac{\pi}{\mathrm{2}}\right)}\:\Rightarrow \\ $$ $${I}\:=\frac{\pi}{\mathrm{2}{cos}\left(\frac{{m}\pi}{\mathrm{2}{k}}\right)} \\ $$ $$\frac{{m}}{\mathrm{2}{k}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\:=\frac{{m}}{\mathrm{2}{k}}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{m}}{{k}}−\mathrm{1}\right)<\mathrm{0}\:\Rightarrow{m}<{k}\:\:{so}\:{the}\:{condition}\:{is} \\ $$ $$\mathrm{1}<{m}<{k} \\ $$

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