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Question Number 84577 by msup trace by abdo last updated on 14/Mar/20

$$calculate\int \frac{dx}{cosx+cos\left(2x\right)+cos\left(3x\right)}$$

Commented by jagoll last updated on 14/Mar/20

$$\cos x+\cos 3x+\cos 2x=$$$$2\cos 2x\cos x+\cos 2x=$$$$\cos 2x(2\cos x+1)$$$$\int \frac{dx}{\cos 2x(2\cos x+1)}=$$$$\mathrm{let}\tan \left(\frac{x}{2}\right)=t$$

Commented by abdomathmax last updated on 14/Mar/20

$$wehavecosx+cos\left(2x\right)+cos\left(3x\right)$$$$=cosx+cos\left(3x\right)+cos\left(2x\right)=2cos\left(2x\right)cosx+cosx$$$$=cosx(2cos\left(2x\right)+1)=cosx(2(2{cos}^{2}x-1)-1)$$$$=cosx(4{cos}^{2}x-3)\Rightarrow$$$$I=\int \frac{dx}{cosx(4{cos}^{2}x-1)}letdecompose$$$$f\left(t\right)=\frac{1}{t(4t^2-1)}=\frac{1}{t(2t-1)(2t+1)}\Rightarrow f\left(t\right)=\frac{a}{t}+\frac{b}{2t-1}+\frac{c}{2t+1}$$$$a=-1$$$$b=\frac{2}{2}=1$$$$c=\frac{-2}{-2}=1\Rightarrow f\left(t\right)=-\frac{1}{t}+\frac{1}{2t-1}+\frac{1}{2t+1}\Rightarrow$$$$I=-\int \frac{dx}{cosx}+\int \frac{dx}{2cosx-1}+\int \frac{dx}{2cosx+1}$$$$\int \frac{dx}{cosx}{=}_{tan\left(\frac{x}{2}\right)=u}\int \frac{2du}{(1+u^2)\times \frac{1-u^2}{1+u^2}}$$$$=\int \frac{2du}{(1-u)(1+u)}=\int (\frac{1}{1-u}+\frac{1}{1+u})du$$$$=ln\mid \frac{1+u}{1-u}\mid +\underset{0}{c}=ln\mid \frac{1+tan\left(\frac{x}{2}\right)}{1-tan\left(\frac{x}{2}\right)}\mid =ln\mid tan(\frac{x}{2}+\frac{\pi }{4})\mid +c_0$$$$\int \frac{dx}{2cosx-1}{=}_{tan\left(\frac{x}{2}\right)=u}\int \frac{2du}{(1+u^2)(2\frac{1-u^2}{1+u^2}-1)}$$$$=\int \frac{2du}{2-2u^2-1-u^2}=\int \frac{2du}{-3u^2+1}$$$$=-2\int \frac{du}{3u^2-1}=-2\int \frac{du}{(\sqrt{3}u+1)(\sqrt{3}u-1)}$$$$=\int (\frac{1}{\sqrt{3}u+1}-\frac{1}{\sqrt{3}u-1})du$$$$=\frac{1}{\sqrt{3}}ln\mid \frac{\sqrt{3}u+1}{\sqrt{3}u-1}\mid +c_1=\frac{1}{\sqrt{3}}ln\mid \frac{\sqrt{3}tan\left(\frac{x}{2}\right)+1}{\sqrt{3}tan\left(\frac{x}{2}\right)-1}\mid +c_1$$$$wefolowthesamemannertodtermine$$$$\int \frac{dx}{2cosx+1}$$

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