Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 211871 by Spillover last updated on 22/Sep/24

Answered by Ghisom last updated on 23/Sep/24

$$=-\underset{0}{\stackrel{\arccos \frac{\sqrt{6}}{3}}{\int }}\frac{2+\tan x}{3-{2\cos }^{2}x-{\sin }^{2}x}dx=$$$$[t=\sqrt{2}\tan x]$$$$=-\frac{1}{2}\underset{0}{\stackrel{1}{\int }}\frac{t+2\sqrt{2}}{t^2+1}dt=$$$$=-{[\sqrt{2}\arctan t+\frac{1}{4}\ln (t^2+1)]}_0^1=$$$$=-\frac{1}{4}(\sqrt{2}\pi +\ln 2)$$

Commented by Spillover last updated on 23/Sep/24

$$correct.thanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com