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Question Number 60893 by mathsolverby Abdo last updated on 26/May/19

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{2}{cos}^{2}x+\sqrt{3}{sin}^{2}x}$$

Commented by maxmathsup by imad last updated on 27/May/19

$$letA={\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{2}{cos}^{2}x+\sqrt{3}{sin}^{2}x}\Rightarrow A={\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{2}\frac{1+cos\left(2x\right)}{2}+\sqrt{3}\frac{1-cos\left(2x\right)}{2}}$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{2}+\sqrt{3}+(\sqrt{2}-\sqrt{3})cos\left(2x\right)}{=}_{tanx=t}2{\int }_0^{+\mathrm{\infty }}\frac{1}{\sqrt{2}+\sqrt{3}+(\sqrt{2}-\sqrt{3})\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{1}{(1+t^2)(\sqrt{2}+\sqrt{3})+\sqrt{2}-\sqrt{3}}dt$$$$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{\sqrt{2}+\sqrt{3}+(\sqrt{2}+\sqrt{3})t^2+\sqrt{2}-\sqrt{3}}=2{\int }_0^{\mathrm{\infty }}\frac{dt}{2\sqrt{2}+(\sqrt{2}+\sqrt{3})t^2}$$$$=\frac{1}{\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+\frac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}}{t}^2}changement\sqrt{\frac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}}}t=ugive$$$$A=\frac{1}{\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}\sqrt{\frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}}du=\frac{\sqrt{\sqrt{2}}}{\sqrt{\sqrt{2}+\sqrt{3}}}{\int }_0^{\mathrm{\infty }}\frac{du}{1+u^2}=\frac{\pi }{2}\frac{\sqrt{\sqrt{2}}}{\sqrt{\sqrt{2}+\sqrt{3}}}.$$

Answered by Prithwish sen last updated on 27/May/19

$$\int \frac{\mathrm{dx}}{\sqrt{2}{\cos }^2\mathrm{x}+\sqrt{3}{\sin }^2\mathrm{x}}=\int \frac{{\sec }^2\mathrm{x}\mathrm{dx}}{\sqrt{2}+\sqrt{3}{\tan }^2\mathrm{x}}$$$$\mathrm{Now}\mathrm{putting}\mathrm{tanx}=\mathrm{v}$$$$\mathrm{then}{\sec }^2\mathrm{x}\mathrm{dx}=\mathrm{dv}$$$$\mathrm{the}\mathrm{integral}\mathrm{changes}\mathrm{to}$$$$\int \frac{\mathrm{dv}}{\sqrt{2}+\sqrt{3}{\mathrm{v}}^2}$$

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