# Question and Answers Forum

Integration Questions

Question Number 185880 by cortano1 last updated on 29/Jan/23

$$\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}{\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:{dx}=? \\$$

Commented by MJS_new last updated on 29/Jan/23

$$\mathrm{simply}\:\mathrm{use}\:{t}=\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}} \\$$

Commented by cortano1 last updated on 29/Jan/23

Answered by cortano1 last updated on 29/Jan/23

$$\:{I}=\:\int\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}{\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:{dx} \\$$$$\:\:=\:\int\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}\:\mathrm{sec}\:^{\mathrm{2}} {x}}{\left(\mathrm{tan}\:{x}+\mathrm{1}\right)^{\mathrm{2}} }\:{dx} \\$$$$\:\:=\:\int\:\frac{{u}^{\mathrm{1}/\mathrm{3}} }{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }\:{du}\: \\$$

Commented by cortano1 last updated on 29/Jan/23

$${it}'{s}\:{correct}? \\$$

Answered by MJS_new last updated on 29/Jan/23

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}{\left(\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)}{dx}= \\$$$$\:\:\:\:\:\left[{t}=\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{3cos}^{\mathrm{2}} \:{x}\:\sqrt[{\mathrm{3}}]{\mathrm{tan}^{\mathrm{2}} \:{x}}{dx}\right] \\$$$$=\mathrm{3}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}^{\mathrm{3}} }{\left({t}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\$$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\$$$$=\left[−\frac{{t}}{{t}^{\mathrm{3}} +\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} +\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{3}} +\mathrm{1}}= \\$$$$=\mathrm{0}+\left[\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} −{t}+\mathrm{1}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}−\mathrm{1}\right)}{\mathrm{3}}\right]_{\mathrm{0}} ^{\infty} = \\$$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\pi \\$$

Commented by cortano1 last updated on 30/Jan/23

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