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Question Number 50421 by Abdo msup. last updated on 16/Dec/18

$$calculateA={\int }_0^{\frac{\pi }{3}}\frac{du}{{(1+{cos}^{2}u)}^3}$$

Commented by maxmathsup by imad last updated on 18/Dec/18

$$wehave1+{cos}^{2}u=1+\frac{1}{1+{tan}^{2}u}=\frac{2+{tan}^{2}u}{1+{tan}^{2}u}\Rightarrow$$$$A={\int }_0^{\frac{\pi }{3}}\frac{du}{{\left(\frac{2+{tan}^{2}u}{1+{tan}^{2}u}\right)}^3}={\int }_0^{\frac{\pi }{3}}\frac{{(1+{tan}^{2}u)}^3}{{(2+{tan}^{2}u)}^3}changementtan\left(u\right)=xgive$$$$A={\int }_0^{\sqrt{3}}\frac{{(1+x^2)}^3}{{(2+x^2)}^3}\frac{dx}{(1+x^2)}={\int }_0^{\sqrt{3}}\frac{x^4+2x^2+1}{{(x^2+2)}^3}dxletdecompose$$$$F\left(x\right)=\frac{x^4+2x^2+1}{{(x^2+2)}^3}$$$$F\left(x\right)=G\left(u\right)=\frac{u^2+2u+1}{{(u+2)}^3}=\frac{a}{u+2}+\frac{b}{{(u+2)}^2}+\frac{c}{{(u+2)}^3}$$$$c={lim}_{u\to -2}{(u+2)}^{2}G\left(u\right)=1$$$${lim}_{u\to +\mathrm{\infty }}uG\left(u\right)=1=a\Rightarrow G\left(u\right)=\frac{1}{u+2}+\frac{b}{{(u+2)}^2}+\frac{1}{{(u+2)}^3}$$$$G(-1)=0=1+b+1=b+2\Rightarrow b=-2\Rightarrow G\left(u\right)=\frac{1}{u+2}-\frac{2}{{(u+2)}^2}+\frac{1}{{(u+2)}^3}\Rightarrow$$$$F\left(x\right)=\frac{1}{x^2+2}-\frac{2}{{(x^2+2)}^2}+\frac{1}{{(x^2+2)}^3}\Rightarrow$$$${\int }_0^{\sqrt{3}}F\left(x\right)dx={\int }_0^{\sqrt{3}}\frac{dx}{x^2+2}-2{\int }_0^{\sqrt{3}}\frac{dx}{{(x^2+2)}^2}+{\int }_0^{\sqrt{3}}\frac{dx}{{(x^2+2)}^3}$$$${\int }_0^{\sqrt{3}}\frac{dx}{x^2+2}{=}_{x=\sqrt{2}u}{\int }_0^{\frac{\sqrt{3}}{\sqrt{2}}}\frac{\sqrt{2}du}{2(1+u^2)}=\frac{1}{\sqrt{2}}arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right).$$$${\int }_0^{\sqrt{3}}\frac{dx}{{(x^2+2)}^2}{=}_{x=\sqrt{2}tan\theta }{\int }_0^{arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right)}\frac{\sqrt{2}(1+{tan}^2\theta )d\theta }{4{(1+{tan}^2\theta )}^2}$$$$=\frac{\sqrt{2}}{4}{\int }_0^{arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right)}{cos}^2\theta d\theta =\frac{\sqrt{2}}{4}{\int }_0^{arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right)}\frac{1+cos\left(2\theta \right)}{2}d\theta$$$$=\frac{\sqrt{2}}{8}arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right)+\frac{\sqrt{2}}{16}{\left[sin\left(2\theta \right)\right]}_0^{arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right)}$$$$=\frac{\sqrt{2}}{8}arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right)+\frac{\sqrt{2}}{16}sin\left(2arctan\left(\frac{\sqrt{3}}{\sqrt{2}}\right)\right)....becontinued...$$

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