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Question Number 27186 by abdo imad last updated on 02/Jan/18

$$findI={\int }_0^{\pi }\frac{dx}{cosx+2sinx}.$$

Commented by abdo imad last updated on 04/Jan/18

$$wedothechangementtan\left(\frac{x}{2}\right)=t$$$$I={\int }_0^{\mathrm{\infty }}\frac{\frac{2dt}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{4t}{1+t^2}}={\int }_0^{\mathrm{\infty }}\frac{2dt}{-t^2+4t+1}$$$$=-2{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2-4t-1}=-2{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2-4t+4-5}$$$$=-2{\int }_0^{\mathrm{\infty }}\frac{dt}{{(t-2)}^2-5}=-2{\int }_0^{\mathrm{\infty }}\frac{dt}{(t-2+\sqrt{5})(t-2-\sqrt{5)}}$$$$=\frac{1}{\sqrt{5}}{\int }_0^{\mathrm{\infty }}(\frac{1}{t-2+\sqrt{}5}-\frac{1}{t-2-\sqrt{5}})$$$$=\frac{1}{\sqrt{5}}{\left[ln/\frac{t-2+\sqrt{5}}{t-2-\sqrt{5}}/\right]}_0^{\propto }$$$$=\frac{1}{\sqrt{5}}(-ln/\frac{-2+\sqrt{5}}{-2-\sqrt{5}}/)=\frac{1}{\sqrt{5}}ln\left(\frac{2+\sqrt{5}}{-2+\sqrt{5}}\right).$$

Answered by mrW1 last updated on 03/Jan/18

$$I={\int }_0^{\pi }\frac{dx}{cosx+2sinx}$$$$=\frac{1}{\sqrt{5}}{\int }_0^{\pi }\frac{dx}{\frac{1}{\sqrt{5}}cosx+\frac{2}{\sqrt{5}}sinx}$$$$=\frac{1}{\sqrt{5}}{\int }_0^{\pi }\frac{dx}{\sin \alpha cosx+\cos \alpha sinx}$$$$=\frac{1}{\sqrt{5}}{\int }_0^{\pi }\frac{dx}{sin(x+\alpha )}$$$$......$$$$=\frac{1}{\sqrt{5}}\ln (9+4\sqrt{5})$$

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