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Question Number 47059 by maxmathsup by imad last updated on 04/Nov/18

$$find\int \frac{dx}{1+cosx+cos\left(2x\right)}$$

Commented by maxmathsup by imad last updated on 06/Nov/18

$$letA=\int \frac{dx}{1+cosx+cos\left(2x\right)}=\int \frac{dx}{1+cos\left(x\right)+2{cos}^{2}x-1}$$$$=\int \frac{dx}{cosx(1+2cosx)}=\int (\frac{1}{cosx}-\frac{2}{1+2cosx})dx=\int \frac{dx}{cosx}-\int \frac{2}{1+2cosx}dx$$$$but\int \frac{dx}{cosx}{=}_{tan\left(\frac{x}{2}\right)=t}\int \frac{1}{\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=\int \frac{2dt}{1-t^2}=\int (\frac{1}{1-t}+\frac{1}{1+t})dt$$$$=ln\mid \frac{1+t}{1-t}\mid +c_1=ln\mid \frac{1+tan\left(\frac{x}{2}\right)}{1-tan\left(\frac{x}{2}\right)}\mid =ln\mid tan(\frac{x}{2}+\frac{\pi }{4})\mid +c_{1}alsowehave$$$$\int \frac{2}{1+2cosx}dt{=}_{tan\left(\frac{x}{2}\right)=t}\int \frac{2}{1+2\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=4\int \frac{dt}{1+t^2+2-2t^2}$$$$=4\int \frac{dt}{3-t^2}{=}_{t=\sqrt{3}u}4\int \frac{\sqrt{3}du}{3(1-u^2)}=\frac{2\sqrt{3}}{4}\int \frac{2du}{1-u^2}=\frac{\sqrt{3}}{2}ln\mid \frac{1+u}{1-u}\mid +c_2$$$$=\frac{\sqrt{3}}{2}ln\mid \frac{1+\frac{t}{\sqrt{3}}}{1-\frac{t}{\sqrt{3}}}\mid +c_2=\frac{\sqrt{3}}{2}ln\mid \frac{\sqrt{3}+t}{\sqrt{3}-t}\mid +c_{2}finally$$$$A=ln\mid tan(\frac{x}{2}+\frac{\pi }{4})\mid -\frac{\sqrt{3}}{2}ln\mid \frac{\sqrt{3}+tan\left(\frac{x}{2}\right)}{\sqrt{3}-tan\left(\frac{x}{2}\right)}\mid +C.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18

$$\int \frac{dx}{cosx+2{cos}^{2}x}$$$$\int \frac{dx}{cosx(1+2cosx)}$$$$\int \frac{1+2cosx-2cosx}{cosx(1+2cosx)}dx$$$$\int \frac{dx}{cosx}-\int \frac{2dx}{2(\frac{1}{2}+cosx)}$$$$\int secxdx-\int \frac{dx}{\frac{1}{2}+\frac{1-{tan}^2\frac{x}{2}}{1+{tan}^2\frac{x}{2}}}$$$$\int secxdx-\int \frac{{sec}^2\frac{x}{2}}{\frac{3}{2}-\frac{1}{2}{tan}^2\frac{x}{2}}dx$$$$\int secxdx-2\int \frac{{sec}^2\frac{x}{2}}{3-{tan}^2\frac{x}{2}}dx$$$$I_1-2I_2$$$$I_1=\int secxdx$$$$=ln\mid sex+tanx\mid +c_1$$$$I_2=\int \frac{{sec}^2\frac{x}{2}}{3-{tan}^2\frac{x}{2}}dx$$$$t=tan\frac{x}{2}dt={sec}^2\frac{x}{2}\times \frac{1}{2}dx$$$$I_2=\int \frac{2dt}{{\left(\sqrt{3}\right)}^2-t^2}$$$$=2\times \frac{1}{2\sqrt{3}}ln\mid \frac{\sqrt{3}+t}{\sqrt{3}-t}\mid +c_2$$$$=\frac{1}{\sqrt{3}}ln\mid \frac{\sqrt{3}+tan\frac{x}{2}}{\sqrt{3}-tan\frac{x}{2}}\mid +c_2$$$$$$

Commented by maxmathsup by imad last updated on 05/Nov/18

$$thankssir.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

$$thankyousir...$$

Answered by ajfour last updated on 04/Nov/18

$$I=\int \frac{dx}{2\cos {}^{2}x+\cos x}$$$$let\tan \frac{x}{2}=t\Rightarrow dx=\frac{2dt}{1+t^2}$$$$I=\int \frac{2dt}{(1+t^2)[2{\left(\frac{1-t^2}{1+t^2}\right)}^2+\frac{1-t^2}{1+t^2}]}$$$$=\int \frac{2(1+t^2)dt}{2{(1-t^2)}^2+1-t^4}$$$$=2\int \frac{(1+t^2)dt}{t^4-4t^2+3}=2\int \frac{(\frac{1}{t^2}+1)dt}{t^2+\frac{3}{t^2}-4}$$$$leta(1-\frac{\sqrt{3}}{t^2})+b(1+\frac{\sqrt{3}}{t^2})=\frac{1}{t^2}+1$$$$\Rightarrow a+b=1\mathrm{\&}b-a=\frac{1}{\sqrt{3}}$$$$\Rightarrow a=\frac{1}{2}-\frac{1}{2\sqrt{3}};b=\frac{1}{2}+\frac{1}{2\sqrt{3}}$$$$I=(1-\frac{1}{\sqrt{3}})\int \frac{(1-\frac{\sqrt{3}}{t^2})dt}{{(t+\frac{\sqrt{3}}{t})}^2-4-2\sqrt{3}}$$$$+(1+\frac{1}{\sqrt{3}})\int \frac{(1+\frac{\sqrt{3}}{t^2})dt}{{(t-\frac{\sqrt{3}}{t})}^2-4+2\sqrt{3}}$$$$I=\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)\frac{1}{2\sqrt{4+2\sqrt{3}}}\ln \mid \frac{t+\frac{\sqrt{3}}{t}-\sqrt{4+2\sqrt{3}}}{t+\frac{\sqrt{3}}{t}+\sqrt{4+2\sqrt{3}}}\mid$$$$+\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)\frac{1}{2\sqrt{4-2\sqrt{3}}}\ln \mid \frac{t+\frac{\sqrt{3}}{t}-\sqrt{4-2\sqrt{3}}}{t+\frac{\sqrt{3}}{t}+\sqrt{4-2\sqrt{3}}}\mid +c.$$

Commented by maxmathsup by imad last updated on 04/Nov/18

$$thankyousirAjfour$$

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