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Question Number 151425 by peter frank last updated on 21/Aug/21

$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{{(\cos \mathrm{x}+\sqrt{3}\sin \mathrm{x})}^2}\mathrm{dx}=\frac{1}{\sqrt{3}}$$

Answered by Olaf_Thorendsen last updated on 21/Aug/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{dx}{{(\cos x+\sqrt{3}\sin x)}^2}$$$$\mathrm{I}=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\frac{dx}{{(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)}^2}$$$$\mathrm{I}=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\frac{dx}{{(\cos \frac{\pi }{3}\cos x+\sin \frac{\pi }{3}\sin x)}^2}$$$$\mathrm{I}=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\frac{dx}{{\cos }^2(x-\frac{\pi }{3})}$$$$\mathrm{I}=\frac{1}{4}{\int }_{-\frac{\pi }{3}}^{\frac{\pi }{6}}\frac{du}{{\cos }^{2}u}$$$$\mathrm{I}=\frac{1}{4}{\left[\tan u\right]}_{-\frac{\pi }{3}}^{\frac{\pi }{6}}$$$$\mathrm{I}=\frac{1}{4}(\frac{1}{\sqrt{3}}+\sqrt{3})=\frac{1}{\sqrt{3}}$$

Answered by MJS_new last updated on 21/Aug/21

$$\int \frac{dx}{{(\cos x+\sqrt{3}\sin x)}^2}=$$$$[t=\tan x\to dx={\cos }^{2}xdt]$$$$=\int \frac{dt}{{(\sqrt{3}t+1)}^2}=-\frac{1}{3t+\sqrt{3}}=\frac{1}{\sqrt{3}+3\tan x}+C$$$${\left[\frac{1}{\sqrt{3}+3\tan x}\right]}_0^{\pi /2}=\frac{\sqrt{3}}{3}$$

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