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Question Number 43490 by peter frank last updated on 11/Sep/18

$$\mathbf{evaluate}$$$$\int \sqrt{\mathbf{tan}\mathit{\theta }\mathbf{d}\theta }$$

Commented by maxmathsup by imad last updated on 11/Sep/18

$$letI=\int \sqrt{tan\theta }d\theta changement\sqrt{tan\theta }=xgivetan\theta =x^2\nLeftrightarrow \theta =arctan\left(x^2\right)\Rightarrow$$$$I=\int x\frac{2x}{1+x^4}dx=\int \frac{2x^2}{1+x^4}dxletdecomposeF\left(x\right)=\frac{2x^2}{1+x^4}$$$$F\left(x\right)=\frac{2x^2}{{(x^2+1)}^2-2x^2}=\frac{2x^2}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}$$$$=\frac{ax+b}{x^2+\sqrt{2}x+1}+\frac{cx+d}{x^2-\sqrt{2}x+1}$$$$F(-x)=F\left(x\right)\Rightarrow \frac{-ax+b}{x^2-\sqrt{2}x+1}+\frac{-cx+d}{x^2+\sqrt{2}x+1}=F\left(x\right)\Rightarrow c=-aandb=d\Rightarrow$$$$F\left(x\right)=\frac{ax+b}{x^2+\sqrt{2}x+1}+\frac{-ax+b}{x^2-\sqrt{2}x+1}$$$$F\left(0\right)=0=2b\Rightarrow b=0\Rightarrow F\left(x\right)=\frac{ax}{x^2+\sqrt{2}x+1}-\frac{ax}{x^2-\sqrt{2}x+1}$$$$F\left(1\right)=1=\frac{a}{2+\sqrt{2}}-\frac{a}{2-\sqrt{2}}\Rightarrow \left(\frac{-2\sqrt{2}}{2}\right)a=1\Rightarrow a=-\frac{1}{\sqrt{2}}\Rightarrow$$$$F\left(x\right)=\frac{1}{\sqrt{2}}\{\frac{1}{x^2-\sqrt{2}x+1}-\frac{1}{x^2+\sqrt{2}x+1}\}\Rightarrow$$$$I=\frac{1}{\sqrt{2}}\int \frac{dx}{x^2-\sqrt{2}x+1}-\frac{1}{\sqrt{2}}\int \frac{dx}{x^2+\sqrt{2}x+1}but$$$$\int \frac{dx}{x^2-\sqrt{2}x+1}=\int \frac{dx}{x^2-\frac{2}{\sqrt{2}}x+\frac{1}{2}+\frac{1}{2}}=\int \frac{dx}{{(x-\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}$$$${=}_{x-\frac{1}{\sqrt{2}}=\frac{u}{\sqrt{2}}}\int \frac{1}{\frac{1}{2}(1+u^2)}\frac{du}{\sqrt{2}}=\sqrt{2}\int \frac{du}{1+u^2}=\sqrt{2}arctan\left(u\right)+c_1$$$$=\sqrt{2}arctan(x\sqrt{2}-1)+c_1=\sqrt{2}arctan(\sqrt{2tan\theta }-1)+c_{1}alsowehave$$$$\int \frac{dx}{x^2+\sqrt{2}x+1}=\int \frac{dx}{{(x+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}{=}_{x+\frac{1}{\sqrt{2}}=\frac{u}{\sqrt{2}}}\int \frac{1}{\frac{1}{2}(1+u^2)}\frac{du}{\sqrt{2}}$$$$=\sqrt{2}arctan(\sqrt{2}x+1)=\sqrt{2}arctan(\sqrt{2tan\theta }+1)\Rightarrow$$$$\int \sqrt{tan\theta }d\theta =arctan(\sqrt{2tan\theta }-1)-arctan(\sqrt{2tan\theta }+1)+c$$

Commented by peter frank last updated on 11/Sep/18

$$thanks$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

$$t^2=tan\theta$$$$2tdt={sec}^2\theta d\theta$$$$\frac{2t}{1+t^4}dt=d\theta$$$$\int \frac{2t}{1+t^4}\times tdt$$$$\int \frac{2t^2}{1+t^4}dt$$$$\int \frac{2}{t^2+\frac{1}{t^2}}dt$$$$\int \frac{1-\frac{1}{t^2}+1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$$$$\int \frac{1-\frac{1}{t^2}}{{(t+\frac{1}{t})}^2-{\left(\sqrt{2}\right)}^2}dt+\int \frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+{\left(\sqrt{2}\right)}^2}dt$$$$\int \frac{d(t+\frac{1}{t})}{{(t+\frac{1}{t})}^2-{\left(\sqrt{2}\right)}^2}+\int \frac{d(t-\frac{1}{t})}{{(t-\frac{1}{t})}^2+{\left(\sqrt{2}\right)}^2}$$$$\frac{1}{2\sqrt{2}}ln\mid \frac{(t+\frac{1}{t})-\sqrt{2}}{(t+\frac{1}{t})+\sqrt{2}}\mid +\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+c$$$$\frac{1}{2\sqrt{2}}ln\mid \frac{\sqrt{tan\theta }+\frac{1}{\sqrt{tan\theta }}-\sqrt{2}}{\sqrt{tan\theta }+\frac{1}{\sqrt{tan\theta }}+\sqrt{2}}\mid +\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{\sqrt{tan\theta }-\frac{1}{\sqrt{tan\theta }}}{\sqrt{2}}\right)+c$$$$$$

Commented by peter frank last updated on 11/Sep/18

$$thanks$$

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