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Question Number 93249 by 1549442205 last updated on 12/May/20

$$calculate{\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{dx}{2+cos2x}$$

Commented by 1549442205 last updated on 12/May/20

$$Calculateitsvalue?$$

Commented by abdomathmax last updated on 12/May/20

$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{2+cos\left(2x\right)}\Rightarrow I{=}_{2x=t}\frac{1}{2}{\int }_0^{\pi }\frac{dt}{2+cost}$$$${=}_{tan\left(\frac{t}{2}\right)=u}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{2du}{(1+u^2)(2+\frac{1-u^2}{1+u^2})}$$$$={\int }_0^{\mathrm{\infty }}\frac{du}{2+2u^2+1-u^2}={\int }_0^{\mathrm{\infty }}\frac{du}{3+u^2}{=}_{u=\sqrt{3}z}$$$$={\int }_0^{\mathrm{\infty }}\frac{\sqrt{3}dz}{3(1+z^2)}=\frac{1}{\sqrt{3}}{\left[arctanz\right]}_0^{+\mathrm{\infty }}=\frac{1}{\sqrt{3}}\times \frac{\pi }{2}\Rightarrow$$$$I=\frac{\pi }{2\sqrt{3}}$$$$$$

Commented by 1549442205 last updated on 12/May/20

$$Thankyouverymuch!$$

Commented by mathmax by abdo last updated on 12/May/20

$$youarewelcome.$$

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