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Question Number 20466 by tammi last updated on 27/Aug/17

∫((sin xcos xdx)/(sin^4 x+cos^4 x))

$$\int\frac{\mathrm{sin}\:{x}\mathrm{cos}\:{xdx}}{\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}} \\ $$

Answered by sma3l2996 last updated on 27/Aug/17

t=sin^2 x⇒dt=2sinxcosxdx I=(1/2)∫(dt/(t^2 +(1−t)^2 ))=(1/2)∫(dt/(2t^2 −2t+1)) =(1/2)∫(dt/(2t^2 −2×(((√2)t)/(√2))+(1/2)−(1/2)+1)) =(1/2)∫(dt/(((√2)t−(1/(√2)))^2 +(1/2))) =(1/2)∫(dt/((((2t−1)/(√2)))^2 +(1/2)))=(1/2)∫(dt/((1/2)((((2t−1)/(√2))×(√2))^2 +1))) =∫(dt/((2t−1)^2 +1)) u=2t−1⇒du=2dt I=(1/2)∫(du/(u^2 +1))=(1/2)tan^(−1) (u)+C=(1/2)tan^(−1) (2t−1)+C =(1/2)tan^(−1) (2sin^2 x−1)+C=(1/2)tan^(−1) (−cos(2x))+C I=−(1/2)tan^(−1) (cos(2x))+C

$${t}={sin}^{\mathrm{2}} {x}\Rightarrow{dt}=\mathrm{2}{sinxcosxdx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}×\frac{\sqrt{\mathrm{2}}{t}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\left(\sqrt{\mathrm{2}}{t}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{2}}}×\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\int\frac{{dt}}{\left(\mathrm{2}{t}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${u}=\mathrm{2}{t}−\mathrm{1}\Rightarrow{du}=\mathrm{2}{dt} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({u}\right)+{C}=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}{t}−\mathrm{1}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}{sin}^{\mathrm{2}} {x}−\mathrm{1}\right)+{C}=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(−{cos}\left(\mathrm{2}{x}\right)\right)+{C} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({cos}\left(\mathrm{2}{x}\right)\right)+{C} \\ $$

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