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Question Number 48104 by wasim last updated on 19/Nov/18

$$\mathrm{solve}\mathrm{this}$$$$\int (2\mathrm{sinx}+\mathrm{cosx})/(2+3\mathrm{sinx}+{\sin }^{2\mathrm{x}})\mathrm{dx}$$

Answered by MJS last updated on 19/Nov/18

$$\mathrm{Weierstrass}-\mathrm{substitution}$$$$t=\tan \frac{x}{2}\Rightarrow x=2\arctan t;dx=\frac{2dt}{1+t^2}$$$$\sin x=\frac{2t}{1+t^2};\cos x=\frac{1-t^2}{1+t^2}$$$$-\int \frac{t^2-4t-1}{{(t+1)}^2(t^2+t+1)}dt=$$$$=2\int \frac{dt}{t+1}-4\int \frac{dt}{{(t+1)}^2}-\int \frac{2t-3}{t^2+t+1}dt=$$$$=2\int \frac{dt}{t+1}-4\int \frac{dt}{{(t+1)}^2}-\int \frac{2x+1}{t^2+t+1}dt+4\int \frac{dt}{t^2+t+1}$$$$=2\ln (t+1)+\frac{4}{t+1}-\ln (t^2+t+1)+\frac{8\sqrt{3}}{3}\arctan \left(\frac{\sqrt{3}}{3}(2t+1)\right)=$$$$=\ln \frac{{(t+1)}^2}{t^2+t+1}+\frac{4}{t+1}+\frac{8\sqrt{3}}{3}\arctan \left(\frac{\sqrt{3}}{3}(2t+1)\right)=$$$$=\ln \frac{1+\sin x}{2+\sin x}+\ln 2+2+2\sec x-2\tan x+\frac{8\sqrt{3}}{3}\arctan \left(\frac{\sqrt{3}}{3}(1+2\tan \frac{x}{2})\right)=$$$$=\ln \frac{1+\sin x}{2+\sin x}+2\sec x-2\tan x+\frac{8\sqrt{3}}{3}\arctan \left(\frac{\sqrt{3}}{3}(1+2\tan \frac{x}{2})\right)+C$$

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