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Question Number 36201 by prof Abdo imad last updated on 30/May/18

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+2{sin}^2\theta }$$

Commented by prof Abdo imad last updated on 01/Jun/18

$$letputI={\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+2{sin}^2\theta }$$$$I={\int }_0^{\frac{\pi }{2}}\frac{d\theta }{1+2\frac{1-cos\left(2\theta \right)}{2}}={\int }_0^{\frac{\pi }{2}}\frac{d\theta }{2-cos\left(2\theta \right)}$$$${=}_{2\theta =t}{\int }_0^{\pi }\frac{1}{2-cos\left(t\right)}\frac{dt}{2}$$$${=}_{tan\left(\frac{t}{2}\right)=x}{\int }_0^{+\mathrm{\infty }}\frac{1}{2\{2-\frac{1-x^2}{1+x^2}\}}\frac{2dx}{1+x^2}$$$$={\int }_0^{+\mathrm{\infty }}\frac{dx}{2(1+x^2)-1+x^2}$$$$={\int }_0^{+\mathrm{\infty }}\frac{dx}{1+3x^2}changementx\sqrt{3}=ugive$$$$I={\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}\frac{du}{\sqrt{3}}=\frac{1}{\sqrt{3}}\frac{\pi }{2}\Rightarrow I=\frac{\pi }{2\sqrt{3}}.$$$$$$

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