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 217122 by efronzo1 last updated on 01/Mar/25

$$\int \frac{\sqrt{\cos 2\mathrm{x}}}{\cos \mathrm{x}}\mathrm{dx}=?$$

Answered by Frix last updated on 01/Mar/25

$$\int \frac{\sqrt{\cos 2x}}{\cos x}dx=\int \frac{\sqrt{-1+{2\cos }^{2}x}}{\cos x}dx\stackrel{[t=\sqrt{2}\sin x]}{=}$$$$=\sqrt{2}\int \frac{\sqrt{1-t^2}}{2-t^2}dt\stackrel{[u=\frac{t}{\sqrt{1-t^2}}]}{=}\sqrt{2}\int \frac{du}{(u^2+1)(u^2+2)}=$$$$=\sqrt{2}\int (\frac{1}{u^2+1}-\frac{1}{u^2+2})du=$$$$=\sqrt{2}{\tan }^{-1}u-{\tan }^{-1}\frac{u}{\sqrt{2}}=$$$$...$$$$=\sqrt{2}{\sin }^{-1}\left(\sqrt{2}\sin x\right)-{\sin }^{-1}\left(\tan x\right)+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com