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Question Number 77103 by Boyka last updated on 03/Jan/20

Commented by turbo msup by abdo last updated on 03/Jan/20

$$letI=\int \frac{2}{2+sinx}dxchangement$$$$tan\left(\frac{x}{2}\right)=tgive$$$$I=\int \frac{2}{2+\frac{2t}{1+t^2}}\times \frac{2dt}{1+t^2}=2\int \frac{dt}{1+t^2+t}$$$$=2\int \frac{dt}{t^2+t+1}=2\int \frac{dt}{{(t+\frac{1}{2})}^2+\frac{3}{4}}$$$${=}_{t+\frac{1}{2}=\frac{\sqrt{3}}{2}u}2\int \frac{1}{\frac{3}{4}(1+u^2)}\times \frac{\sqrt{3}}{2}du$$$$=2\times \frac{4}{3}\times \frac{\sqrt{3}}{2}arctanu+c$$$$=\frac{4}{\sqrt{3}}arctan\left(\frac{2t+1}{\sqrt{3}}\right)+c$$$$=\frac{4}{\sqrt{3}}arctan\left(\frac{2tan\left(\frac{x}{2}\right)+1}{\sqrt{3}}\right)+C$$

Answered by john santu last updated on 03/Jan/20

$$let\tan \left(\frac{x}{2}\right)=u.\Rightarrow \sin \left(x\right)=\frac{2u}{1+u^2}$$$$dx=\frac{2}{1+u^2}du.\Rightarrow 2+\sin \left(x\right)=\frac{2u^2+2u+2}{1+u^2}$$$$\int \frac{4}{2(u^2+u+1)}du=$$$$\int \frac{2du}{{(u+\frac{1}{2})}^2+\frac{3}{4}}=2\times \frac{\sqrt{3}}{2}{\tan }^{-1}(\tan \left(\frac{x}{2}\right)+\frac{\sqrt{3}}{2})+c$$$$=\sqrt{3}{\tan }^{-1}(\tan \left(\frac{x}{2}\right)+\frac{\sqrt{3}}{2})+c★$$

Commented by MJS last updated on 03/Jan/20

$$\mathrm{error}$$$$\mathrm{line}2\mathrm{must}\mathrm{be}2+\sin x=\frac{2(t^2+t+1)}{t^2+1}$$

Commented by john santu last updated on 03/Jan/20

$$ohyessir$$$$$$

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