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Question Number 137055 by mathmax by abdo last updated on 29/Mar/21

$$\mathrm{find}{\int }_0^{2\pi }\frac{\mathrm{dx}}{{(1+\mathrm{cosx}+3\mathrm{sinx})}^2}$$

Commented by MJS_new last updated on 30/Mar/21

$$0⩽{(1+\cos x+3\sin x)}^2⩽11+2\sqrt{10}$$$$\Rightarrow$$$$\frac{11-2\sqrt{10}}{81}⩽\frac{1}{{(1+\cos x+3\sin x)}^2}⩽+\mathrm{\infty }$$$$\Rightarrow$$$$\mathrm{integral}\mathrm{diverges}$$

Answered by Dwaipayan Shikari last updated on 30/Mar/21

$${\int }_0^{2\pi }\frac{dx}{(2{cos}^2\frac{x}{2}+6sin\frac{x}{2}cos\frac{x}{2})}$$$$=\frac{1}{4}\int \frac{{sec}^2\frac{x}{2}}{{(cos\frac{x}{2}+3sin\frac{x}{2})}^2}dx=\frac{1}{4}\int \frac{{sec}^4\frac{x}{2}}{{(1+3tan\frac{x}{2})}^2}dx$$$$=\frac{1}{2}\int \frac{{sec}^{4}u}{{(1+3tanu)}^2}du=\frac{1}{2}\int \frac{(1+t^2)}{{(1+3t)}^2}dt$$$$=\frac{1}{18}\int \frac{dt}{{(t+\frac{1}{3})}^2}+\frac{1}{18}\int \frac{t^2+\frac{2t}{3}+\frac{1}{9}}{{(t+\frac{1}{3})}^2}-\frac{1}{18}\int \frac{\frac{2t}{3}+\frac{1}{9}}{{(t+\frac{1}{3})}^2}dt$$$$=-\frac{1}{18}.\frac{1}{t+\frac{1}{3}}+\frac{1}{18}-\frac{1}{27}log(t+\frac{1}{3})+\frac{1}{162(t+\frac{1}{3})}-\frac{1}{81(t+\frac{1}{3})}$$$$={[\frac{1}{tan\left(\frac{x}{2}\right)}(\frac{1}{162}-\frac{1}{18}-\frac{1}{81})+\frac{1}{18}-\frac{1}{27}log(tan\frac{x}{2}+\frac{1}{3})]}_0^{2\pi }$$$$Diverges$$

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