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Question Number 43914 by maxmathsup by imad last updated on 17/Sep/18

$$find\int \frac{x}{\sqrt{1+cosx}}dx.$$

Commented by maxmathsup by imad last updated on 19/Sep/18

$$letA=\int \frac{x}{\sqrt{1+cosx}}dxwehaveA=\int \frac{x}{\sqrt{2{cos}^2\left(\frac{x}{2}\right)}}dx$$$$=\frac{1}{\sqrt{2}}\int \frac{x}{cos\left(\frac{x}{2}\right)}dx{=}_{\frac{x}{2}=t}\frac{1}{\sqrt{2}}\int \frac{2t}{cost}2dt=2\sqrt{2}\int \frac{t}{cost}dt$$$${=}_{tan\left(\frac{t}{2}\right)=u}2\sqrt{2}\int \frac{2arctanu}{\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}=8\sqrt{2}\int \frac{arctan\left(u\right)}{1-u^2}du$$$$letconsidertheparametricfunctionf\left(\alpha \right)=\int \frac{arctan\left(\alpha u\right)}{1-u^2}du$$$$wehave{f}^{{}^{\prime }}\left(\alpha \right)=\int \frac{u}{(1-u^2)(1+{\alpha }^2{u}^2)}duletdecompose$$$$F\left(u\right)=\frac{u}{(1-u^2)(1+{\alpha }^2{u}^2)}=\frac{a}{1-u}+\frac{b}{1+u}+\frac{cu+d}{{\alpha }^2{u}^2+1}=\frac{u}{(1-u)(1+u)({\alpha }^2{u}^2+1)}$$$$a={lim}_{u\to 1}(1-u)F\left(u\right)=\frac{1}{2(1+{\alpha }^2)}$$$$b={lim}_{u\to -1}(1+u)F\left(u\right)=\frac{-1}{2(1+{\alpha }^2)}\Rightarrow$$$$F\left(u\right)=\frac{1}{2(1+{\alpha }^2)(1-u)}-\frac{1}{2(1+{\alpha }^2)(1+u)}+\frac{cu+d}{{\alpha }^2{u}^2+1}$$$${lim}_{u\to +\mathrm{\infty }}uF\left(u\right)=0=-a+b+\frac{c}{{\alpha }^2}\Rightarrow \frac{c}{{\alpha }^2}=a-b=\frac{1}{2(1+{\alpha }^2)}+\frac{1}{2(1+{\alpha }^2)}$$$$=\frac{1}{1+{\alpha }^2}\Rightarrow c=\frac{{\alpha }^2}{1+{\alpha }^2}\Rightarrow F\left(u\right)=\frac{1}{2(1+{\alpha }^2)(1-u)}-\frac{1}{2(1+{\alpha }^2)(1+u)}+\frac{\frac{{\alpha }^2}{1+{\alpha }^2}u+d}{1+{\alpha }^2{u}^2}$$$$F\left(o\right)=0=d\Rightarrow \int F\left(u\right)du=\frac{1}{2(1+{\alpha }^2)}\int (\frac{1}{1-u}-\frac{1}{1+u})du$$$$+\frac{{\alpha }^2}{1+{\alpha }^2}\int \frac{u}{1+{\alpha }^2{u}^2}dubut\int (\frac{1}{1-u}-\frac{1}{1+u})du=-ln\mid 1-u^2\mid +c_1$$$$\frac{1}{1+{\alpha }^2}\int \frac{{\alpha }^{2}u}{1+{\alpha }^2{u}^2}du=\frac{1}{2(1+{\alpha }^2)}ln\mid 1+{\alpha }^2{u}^2\mid +c_2\Rightarrow$$$$f^{{}^{\prime }}\left(\alpha \right)=-\frac{ln\mid 1-u^2\mid }{2(1+{\alpha }^2)}+\frac{1}{2(1+{\alpha }^2)}ln\mid 1+{\alpha }^2{u}^2\mid +k.....becontinued...$$

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