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Question Number 205645 by hardmath last updated on 26/Mar/24

$$\mathrm{Find}:\mathrm{\Omega }={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{{\sin }^2\mathrm{x}}{2\mathrm{cosx}+3\mathrm{sinx}}\mathrm{dx}=?$$

Answered by Frix last updated on 26/Mar/24

$$t=\tan \frac{x}{2}\mathrm{leads}\mathrm{to}$$$$-4\int \frac{t^2}{{(t^2+1)}^2(t^2-3t-1)}dt$$$$\mathrm{which}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{but}...$$$$\mathrm{I}\mathrm{found}\mathrm{this}:$$$$\frac{s^2}{2c+3s}=\frac{s^2(2c-3s)}{(2c+3s)(2c-3s)}=\frac{3s^3}{13s^2-4}-\frac{2{cs}^2}{13s^2-4}$$$$\Rightarrow$$$${\mathrm{\Omega }}_1=3\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{{\sin }^{3}x}{{13\sin }^{2}x-4}dx\stackrel{u=\cos x}{=}3\underset{0}{\stackrel{1}{\int }}\frac{u^2-1}{13u^2-9}du=$$$$={[\frac{3u}{13}+\frac{2\sqrt{13}}{169}\ln \mid \frac{\sqrt{13}u+3}{\sqrt{13}u-3}\mid ]}_0^1=$$$$\frac{3}{13}+\frac{4\sqrt{13}}{169}(\ln 2-\ln (-3+\sqrt{13}))$$$${\mathrm{\Omega }}_2=-2\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\cos x{\sin }^{2}x}{{13\sin }^{2}x-4}dx\stackrel{v=\sin x}{=}-2\underset{0}{\stackrel{1}{\int }}\frac{v^2}{13v^2-4}dv=$$$$={[-\frac{2v}{13}+\frac{2\sqrt{13}}{169}\ln \mid \frac{\sqrt{13}v+2}{\sqrt{13}v-2}\mid ]}_0^1=$$$$=-\frac{2}{13}+\frac{4\sqrt{13}}{169}(\ln (2+\sqrt{13})-\ln 3))$$$$\Rightarrow$$$$\mathrm{\Omega }=\frac{1}{13}+\frac{4\sqrt{13}}{169}(2\ln (5+\sqrt{13})-\ln 3-2\ln 2)$$$$\approx .232231309$$

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