Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 125711 by TITA last updated on 13/Dec/20

$$\mathrm{I}=\int \frac{\cos 2\mathrm{x}}{1+\cos {}^2\mathrm{x}}\mathrm{dx}=?\mathrm{please}\mathrm{help}$$$$$$

Answered by bramlexs22 last updated on 13/Dec/20

$$let\tan \left(x\right)=t\wedge dt=\frac{dx}{\cos {}^{2}x}$$$$I=\int \frac{2\cos {}^{2}x-1}{1+\cos {}^{2}x}dx=\int \frac{2\left(\frac{1}{1+t^2}\right)-1}{1+\left(\frac{1}{1+t^2}\right)}\left[\frac{1}{1+t^2}\right]dt$$$$I=\int \left(\frac{1-t^2}{2+t^2}\right)\left(\frac{1}{1+t^2}\right)dt$$$$decompositionpartialfraction$$$$\frac{1-t^2}{(t^2+2)(t^2+1)}=\frac{At+B}{t^2+2}+\frac{Ct+D}{t^2+1}$$$$1-t^2=(At+B)(t^2+1)+(Ct+D)(t^2+2)$$$$t=0\Rightarrow 1=B+2D$$$$t=i\Rightarrow 2=Ci+D$$$$t=-i\Rightarrow 2=-Ci+D;givesD=2\wedge C=0$$$$B=-3$$$$t=-1\Rightarrow 0=(-A-3)\left(2\right)+2\left(3\right)$$$$-3=-A-3;A=0$$$$I=\int -\frac{3}{t^2+2}dt+\int \frac{2}{t^2+1}dt$$$$I=-\frac{3}{\sqrt{2}}\arctan \left(\frac{t}{\sqrt{2}}\right)+2\arctan \left(t\right)+c$$$$I=-\frac{3}{\sqrt{2}}\arctan \left(\frac{\tan x}{\sqrt{2}}\right)+2\arctan \left(\tan x\right)+c$$$$I=2x-\frac{3}{\sqrt{2}}\arctan \left(\frac{\tan x}{\sqrt{2}}\right)+c$$$$$$

Answered by MJS_new last updated on 14/Dec/20

$$\frac{\cos 2x}{1+{\cos }^{2}x}=\frac{-1+{2\cos }^{2}x}{1+{\cos }^{2}x}=2-\frac{3}{1+{\cos }^{2}x}$$$$\int \frac{\cos 2x}{1+{\cos }^{2}x}dx=2\int dx-3\int \frac{dx}{1+{\cos }^{2}x}$$$$2\int dx=2x$$$$-3\int \frac{dx}{1+{\cos }^{2}x}=$$$$[t=\tan x\to dx={\cos }^{2}xdt]$$$$=-3\int \frac{dt}{t^2+2}=-\frac{3\sqrt{2}}{2}\arctan \frac{\sqrt{2}t}{2}$$$$\Rightarrow$$$$I=2x-\frac{3\sqrt{2}}{2}\arctan \frac{\sqrt{2}t}{2}+C$$

Commented by bramlexs22 last updated on 14/Dec/20

$$whysir-\frac{3\sqrt{2}}{2}\arctan \frac{\sqrt{2}t}{2}i?$$$$ittypo?$$

Commented by MJS_new last updated on 14/Dec/20

$$\mathrm{yes},\mathrm{sorry}$$

Answered by mathmax by abdo last updated on 14/Dec/20

$$\mathrm{I}=\int \frac{\cos \left(2\mathrm{x}\right)}{1+{\cos }^2\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{I}=\int \frac{\cos \left(2\mathrm{x}\right)}{1+\frac{1+\cos \left(2\mathrm{x}\right)}{2}}\mathrm{dx}=2\int \frac{\cos \left(2\mathrm{x}\right)}{3+\cos \left(2\mathrm{x}\right)}\mathrm{dx}$$$${=}_{2\mathrm{x}=\mathrm{t}}2\int \frac{\mathrm{cost}}{3+\mathrm{cost}}\frac{\mathrm{dt}}{2}=\int \frac{\mathrm{cost}}{3+\mathrm{cost}}\mathrm{dt}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}}$$$$=\int \frac{\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2}}{3+\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2}}\times \frac{2\mathrm{du}}{1+{\mathrm{u}}^2}=\int \frac{1-{\mathrm{u}}^2}{(1+{\mathrm{u}}^2)(3+{3\mathrm{u}}^2+1-{\mathrm{u}}^2)}\mathrm{du}$$$$=\int \frac{1-{\mathrm{u}}^2}{({\mathrm{u}}^2+1)({2\mathrm{u}}^2+4)}\mathrm{du}=\frac{1}{2}\int \frac{1-{\mathrm{u}}^2}{({\mathrm{u}}^2+1)({\mathrm{u}}^2+2)}\mathrm{du}$$$$=\frac{1}{2}\int (1-{\mathrm{u}}^2)(\frac{1}{{\mathrm{u}}^2+1}-\frac{1}{{\mathrm{u}}^2+2})\mathrm{du}$$$$=-\frac{1}{2}\int \frac{{\mathrm{u}}^2-1}{{\mathrm{u}}^2+1}\mathrm{du}+\frac{1}{2}\int \frac{{\mathrm{u}}^2-1}{{\mathrm{u}}^2+2}\mathrm{du}$$$$=-\frac{1}{2}\int \frac{{\mathrm{u}}^2+1-2}{{\mathrm{u}}^2+1}\mathrm{du}+\frac{1}{2}\int \frac{{\mathrm{u}}^2+2-3}{{\mathrm{u}}^2+2}\mathrm{du}$$$$=\int \frac{\mathrm{du}}{{\mathrm{u}}^2+1}-\frac{3}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+2}(\to \mathrm{u}=\sqrt{2}\alpha )$$$$=\arctan \left(\mathrm{u}\right)-\frac{3}{2}\int \frac{\sqrt{2}\mathrm{d}\alpha }{2(1+{\alpha }^2)}=\mathrm{arctanu}-\frac{3\sqrt{2}}{4}\arctan \left(\frac{\mathrm{u}}{\sqrt{2}}\right)+\mathrm{c}$$$$\mathrm{I}=\mathrm{x}-\frac{3\sqrt{2}}{4}\arctan \left(\frac{\mathrm{tanx}}{\sqrt{2}}\right)+\mathrm{C}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com