Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 214341 by liuxinnan last updated on 06/Dec/24

$$\int \frac{dx}{3+cosx}=?$$

Answered by chhaythean last updated on 06/Dec/24

$$\mathrm{let},\mathrm{t}=\tan \frac{x}{2}\Rightarrow \mathrm{dt}=\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)\mathrm{d}x$$$$\mathrm{or}\mathrm{dt}=\frac{1}{2}(1+{\tan }^2\left(\frac{x}{2}\right))\mathrm{d}x=\frac{1}{2}(1+{\mathrm{t}}^2)\mathrm{d}x$$$$\Rightarrow \cos x=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}$$$$\Rightarrow \int \frac{\mathrm{d}x}{3+\cos x}=2\int \frac{1}{{(1+\mathrm{t})}^2}\times \frac{1}{3+\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}\mathrm{dt}$$$$=2\int \frac{1}{(1+{\mathrm{t}}^2)}\times \frac{1}{\frac{3+{3\mathrm{t}}^2+1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}\mathrm{dt}$$$$=2\int \frac{\mathrm{dt}}{4+{2\mathrm{t}}^2}=\int \frac{\mathrm{dt}}{2+{\mathrm{t}}^2}=\int \frac{\mathrm{dt}}{{\left(\sqrt{2}\right)}^2+{\mathrm{t}}^2}=\frac{1}{\sqrt{2}}\arctan \left(\frac{\mathrm{t}}{\sqrt{2}}\right)+C$$$$=\frac{\sqrt{2}}{2}\arctan \left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{2}}\right)+C$$

Answered by shunmisaki007 last updated on 06/Dec/24

$$\mathrm{Let}t=\tan \left(\frac{x}{2}\right)$$$$;\cos \left(\frac{x}{2}\right)=\frac{1}{\sqrt{1+t^2}},\sin \left(\frac{x}{2}\right)=\frac{t}{\sqrt{1+t^2}},$$$$\cos \left(x\right)={\cos }^2\left(\frac{x}{2}\right)-{\sin }^2\left(\frac{x}{2}\right)=\frac{1-t^2}{1+t^2},$$$$dt=\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=\frac{1+t^2}{2}dx\mathrm{or}dx=\frac{2}{1+t^2}dt.$$$$\mathrm{So}\int \frac{dx}{3+\cos \left(x\right)}=\int \frac{\frac{2}{1+t^2}}{3+\frac{1-t^2}{1+t^2}}dt=\int \frac{2}{4+2t^2}dt=\int \frac{1}{2+t^2}dt$$$$\mathrm{Let}t=\sqrt{2}\tan \left(u\right)$$$$;dt=\sqrt{2}{\sec }^2\left(u\right)du.$$$$\mathrm{So}\int \frac{1}{2+t^2}dt=\int \frac{\sqrt{2}{\sec }^2\left(u\right)}{2+{2\tan }^2\left(u\right)}du=\frac{\sqrt{2}}{2}\int du$$$$=\frac{\sqrt{2}}{2}u+C=\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{\sqrt{2}t}{2}\right)+C$$$$=\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{\sqrt{2}\tan \left(\frac{x}{2}\right)}{2}\right)+C$$$$\therefore \int \frac{dx}{3+\cos \left(x\right)}=\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{\sqrt{2}\tan \left(\frac{x}{2}\right)}{2}\right)+C★$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com