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Question Number 32716 by caravan msup abdo. last updated on 31/Mar/18

$$find{\int }_0^{2\pi }\frac{{cos}^{2}x}{1+3{sin}^{2}x}dx.$$

Answered by Joel578 last updated on 02/Apr/18

$$I=\int \frac{\frac{1+\cos 2x}{2}}{1+\frac{3(1-\cos 2x)}{2}}dx=\int \frac{1+\cos 2x}{5-3\cos 2x}dx$$$$=\int \frac{1+\frac{1-t^2}{1+t^2}}{5-\frac{3(1-t^2)}{1+t^2}}\left(\frac{dt}{1+t^2}\right)\to t=\tan x$$$$=\int \frac{2}{2+8t^2}\left(\frac{dt}{1+t^2}\right)=\int \frac{dt}{(1+4t^2)(1+t^2)}$$$$=\int \frac{4}{3}\left(\frac{1}{1+4t^2}\right)-\frac{1}{3}\left(\frac{1}{1+t^2}\right)dt$$$$=\frac{2}{3}{\tan }^{-1}\left(2t\right)-\frac{1}{3}{\tan }^{-1}\left(t\right)$$$$=\frac{2}{3}{\tan }^{-1}\left(2\tan x\right)-\frac{x}{3}$$$$$$$$I=4{[\frac{2}{3}{\tan }^{-1}\left(2\tan x\right)-\frac{x}{3}]}_0^{\frac{\pi }{2}}$$$$=4[(\frac{2}{3}{\tan }^{-1}\left(2\tan \frac{\pi }{2}\right)-\frac{\pi }{6})-0]$$$$=4(\frac{\pi }{3}-\frac{\pi }{6})=\frac{2\pi }{3}$$

Commented by Joel578 last updated on 02/Apr/18

$$\mathrm{My}\mathrm{first}\mathrm{try}.$$$$\mathrm{Please}\mathrm{recheck}\mathrm{my}\mathrm{solution}$$

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