Question Number 218322 by shunmisaki007 last updated on 06/Apr/25
$$\mathrm{Evaluate}\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{sin}\left({x}\right)}{\mathrm{sin}^{\mathrm{3}} \left({x}\right)+\mathrm{cos}^{\mathrm{3}} \left({x}\right)}\:{dx}. \\ $$
Answered by Ghisom last updated on 06/Apr/25
![∫_0 ^(π/3) ((sin x)/(cos^3 x +sin^3 x))dx= [t=tan x] =∫_0 ^∞ (t/(t^3 +1))dt= =(1/2)∫_0 ^∞ (dt/(t^2 −t+1))+(1/6)∫_0 ^∞ ((2t−1)/(t^2 −t+1))dt−(1/3)∫_0 ^∞ (dt/(t+1))= =[(1/( (√3)))arctan ((2t−1)/( (√3))) +(1/6)ln (t^2 −t+1) −(1/3)ln (t+1)]_0 ^∞ = =[(1/( (√3)))arctan ((2t−1)/( (√3))) +(1/6)ln ((t^2 −t+1)/((t+1)^2 ))]_0 ^∞ = =((2(√3)π)/9)](https://www.tinkutara.com/question/Q218329.png)
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{3}} {\int}}\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{3}} \:{x}\:+\mathrm{sin}^{\mathrm{3}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\right] \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}}{{t}^{\mathrm{3}} +\mathrm{1}}{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}−\frac{\mathrm{1}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}+\mathrm{1}}= \\ $$$$=\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\:\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left({t}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\infty} = \\ $$$$=\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\:\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} = \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}\pi}{\mathrm{9}} \\ $$