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Question Number 144064 by 0731619 last updated on 21/Jun/21

Answered by MJS_new last updated on 21/Jun/21

$$\mathrm{simply}\mathrm{let}t=\tan x$$$$\Rightarrow \mathrm{answer}\mathrm{is}$$$$-\frac{x}{b^2}+\frac{\sqrt{a^2+b^2}}{{ab}^2}\arctan \frac{a\tan x}{\sqrt{a^2+b^2}}+C$$

Commented by 0731619 last updated on 21/Jun/21

$$provethis$$

Commented by MJS_new last updated on 21/Jun/21

$$\mathrm{easy}\mathrm{to}\mathrm{prove}$$$$\frac{d}{dx}[-\frac{x}{b^2}]=-\frac{1}{b^2}$$$$\frac{d}{dx}\left[\frac{\sqrt{a^2+b^2}}{{ab}^2}\arctan \frac{a\tan x}{\sqrt{a^2+b^2}}\right]=$$$$=\frac{\sqrt{a^2+b^2}}{{ab}^2}\times \frac{a\sqrt{a^2+b^2}}{a^2+b^2+a^2{\tan }^{2}x}\times \frac{1}{{\cos }^{2}x}=$$$$=\frac{a^2+b^2}{b^2(a^2+b^2{\cos }^{2}x)}$$$$\frac{a^2+b^2}{b^2(a^2+b^2{\cos }^{2}x)}-\frac{1}{b^2}=\frac{{\sin }^{2}x}{a^2+b^2{\cos }^{2}x}q.e.d.$$

Commented by Dwaipayan Shikari last updated on 21/Jun/21

$$Q.E.D$$

Answered by mathmax by abdo last updated on 21/Jun/21

$$\mathrm{\Phi }=\int \frac{{\sin }^2\mathrm{x}}{{\mathrm{b}}^2{\cos }^2\mathrm{x}+{\mathrm{a}}^2}\mathrm{dx}=\int \frac{{\sin }^2\mathrm{xdx}}{{\mathrm{b}}^2({\cos }^2\mathrm{x}+\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2})}=\frac{1}{{\mathrm{b}}^2}\int \frac{{\sin }^2\mathrm{x}}{\lambda +{\cos }^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{with}\lambda =\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}>0$$$$\mathrm{f}\left(\lambda \right)=\int \frac{{\sin }^2\mathrm{x}}{\lambda +{\cos }^2\mathrm{x}}\mathrm{dx}=\int \frac{\frac{1-\cos \left(2\mathrm{x}\right)}{2}}{\lambda +\frac{1+\cos \left(2\mathrm{x}\right)}{2}}\mathrm{dx}$$$$=\int \frac{1-\cos \left(2\mathrm{x}\right)}{2\lambda +1+\cos \left(2\mathrm{x}\right)}\mathrm{dx}{=}_{2\mathrm{x}=\mathrm{t}}\frac{1}{2}\int \frac{1-\mathrm{cost}}{2\lambda +1+\mathrm{cost}}\mathrm{dt}$$$${=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{y}}\frac{1}{2}\int \frac{1-\frac{1-{\mathrm{y}}^2}{1+{\mathrm{y}}^2}}{2\lambda +1+\frac{1-{\mathrm{y}}^2}{1+{\mathrm{y}}^2}}\times \frac{2\mathrm{dy}}{1+{\mathrm{y}}^2}$$$$=\int \frac{{2\mathrm{y}}^2}{(2\lambda +1)+(2\lambda +1){\mathrm{y}}^2+1-{\mathrm{y}}^2)}\frac{\mathrm{dy}}{(1+{\mathrm{y}}^2)}$$$$=2\int \frac{{\mathrm{y}}^2}{(1+{\mathrm{y}}^2)(2\lambda +2+2\lambda {\mathrm{y}}^2)}\mathrm{dy}=\int \frac{{\mathrm{y}}^2}{({\mathrm{y}}^2+1)(\lambda +1+{\mathrm{y}}^2)}\mathrm{dy}$$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{{\mathrm{y}}^2}{({\mathrm{y}}^2+1)({\mathrm{y}}^2+\lambda +1)}=\frac{1}{\lambda }(\frac{{\mathrm{y}}^2}{{\mathrm{y}}^2+1}-\frac{{\mathrm{y}}^2}{{\mathrm{y}}^2+\lambda +1})$$$$=\frac{1}{\lambda }\{\frac{{\mathrm{y}}^2+1-1}{{\mathrm{y}}^2+1}-\frac{{\mathrm{y}}^2+\lambda +1-\lambda -1}{{\mathrm{y}}^2+\lambda +1}\}$$$$=\frac{1}{\lambda }\{-\frac{1}{{\mathrm{y}}^2+1}+\frac{\lambda +1}{{\mathrm{y}}^2+\lambda +1}\}$$

Commented by mathmax by abdo last updated on 21/Jun/21

$$\mathrm{f}\left(\lambda \right)=-\frac{1}{\lambda }\int \frac{\mathrm{dy}}{{\mathrm{y}}^2+1}+\frac{\lambda +1}{\lambda }\int \frac{\mathrm{dy}}{{\mathrm{y}}^2+\lambda +1}(\to \mathrm{y}=\sqrt{\lambda +1}\mathrm{z})$$$$=-\frac{1}{\lambda }\mathrm{arctany}+\frac{\lambda +1}{\lambda }\int \frac{\sqrt{\lambda +1}\mathrm{dz}}{(\lambda +1)(1+{\mathrm{z}}^2)}$$$$=-\frac{1}{\lambda }\mathrm{arctany}+\frac{\sqrt{\lambda +1}}{\lambda }\arctan \left(\frac{\mathrm{y}}{\sqrt{\lambda +1}}\right)+\mathrm{k}$$$$=-\frac{\mathrm{x}}{\lambda }+\frac{\sqrt{\lambda +1}}{\lambda }\arctan \left(\frac{1}{\sqrt{\lambda +1}}\mathrm{tanx}\right)+\mathrm{K}$$$$\mathrm{\Phi }=\frac{1}{{\mathrm{b}}^2}(-\frac{\mathrm{x}}{\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}}+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\sqrt{\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}+1}\arctan \left(\frac{1}{\sqrt{\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}+1}}\mathrm{tanx}\right)+\mathrm{K}$$$$=-\frac{\mathrm{x}}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{a}}^2}\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{b}}^2}}\arctan \left(\frac{\mid \mathrm{b}\mid }{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{tanx}\right)+\mathrm{K}$$

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