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Question Number 53295 by gunawan last updated on 20/Jan/19

$${\int }_0^{\frac{\pi }{2}}\frac{1}{2+\cos x}dx=...$$

Commented by maxmathsup by imad last updated on 20/Jan/19

$$changementtan\left(\frac{x}{2}\right)=tgive$$$$A={\int }_0^{\frac{\pi }{2}}\frac{dx}{2+cosx}={\int }_0^1\frac{1}{2+\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}={\int }_0^1\frac{2dt}{2+2t^2+1-t^2}=2{\int }_0^1\frac{dt}{3+t^2}$$$${=}_{t=\sqrt{3}u}2{\int }_0^{\frac{1}{\sqrt{3}}}\frac{\sqrt{3}du}{3(1+u^2)}=\frac{2\sqrt{3}}{3}{\int }_0^{\frac{1}{\sqrt{3}}}\frac{du}{1+u^2}=\frac{2\sqrt{3}}{3}arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{2}{\sqrt{3}}\frac{\pi }{6}$$$$=\frac{\pi }{3\sqrt{3}}\Rightarrow ★I=\frac{\pi }{3\sqrt{3}}★$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

$${\int }_0^{\frac{\pi }{2}}\frac{1}{2+\frac{1-{tan}^2\frac{x}{2}}{1+{tan}^2\frac{x}{2}}}dx$$$${\int }_0^{\frac{\pi }{2}}\frac{{sec}^2\frac{x}{2}}{2+2{tan}^2\frac{x}{2}+1-{tan}^2\frac{x}{2}}dx$$$$t=tan\frac{x}{2}dt=\frac{1}{2}{sec}^2\frac{x}{2}dx$$$${\int }_0^1\frac{2dt}{3+t^2}$$$$2\times \frac{1}{\sqrt{3}}\mid {tan}^{-1}\left(\frac{t}{\sqrt{3}}\right){\mid }_0^1$$$$\frac{2}{\sqrt{3}}\left[{tan}^{-1}\left(\frac{1}{\sqrt{3}}\right)\right]=\frac{2}{\sqrt{3}}\times \frac{\pi }{6}=\frac{\pi }{3\sqrt{3}}$$

Commented by gunawan last updated on 20/Jan/19

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

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