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Question Number 195254 by Spillover last updated on 28/Jul/23

$$\int \frac{\sin {}^{2}x\cos {}^{2}x}{(\sin {}^{5}x+\cos {}^{3}x\sin {}^{2}x+\sin {}^{3}x\cos {}^{2}x+\cos {}^{5}x)}dx$$

Commented by Frix last updated on 28/Jul/23

$$=\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}dx$$$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{post}\mathrm{my}\mathrm{solution}\mathrm{later}$$

Answered by Frix last updated on 28/Jul/23

$$\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}dx\stackrel{t=\sin x+\cos x}{=}$$$$=\frac{1}{2}\int \frac{{(t^2-1)}^2}{t(3-t^2)\sqrt{2-t^2}}dt{=}_{\mathrm{show}\mathrm{this}\mathrm{in}\mathrm{detail}]}^{[\mathrm{sorry}\mathrm{no}\mathrm{time}\mathrm{to}}$$$$=-\frac{1}{2}\int \frac{t}{\sqrt{2-t^2}}dt+\frac{1}{6}\int \frac{dt}{t\sqrt{2-t^2}}-\frac{2}{3}\int \frac{t}{(t^2-3)\sqrt{2-t^2}}dt=$$$$=\frac{\sqrt{2-t^2}}{2}+\frac{\sqrt{2}}{12}\ln \mid \frac{\sqrt{2-t^2}-\sqrt{2}}{t}\mid -\frac{2}{3}{\tan }^{-1}\sqrt{2-t^2}$$$$\mathrm{Now}\mathrm{insert}t=\sin x+\cos x=\sqrt{2}\sin (x+\frac{\pi }{4})$$

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