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 113656 by bobhans last updated on 14/Sep/20

$$\int \frac{{(1+\tan \left(\frac{3\mathrm{x}}{2}\right))}^2}{1+\sin 3\mathrm{x}}\mathrm{dx}?$$

Answered by john santu last updated on 14/Sep/20

$$setting\tan \left(\frac{3x}{2}\right)=s\to \sin 3x=\frac{2s}{1+s^2}$$$$1+\sin 3x=\frac{{(1+s)}^2}{1+s^2}$$$$\frac{3}{2}\sec {}^2\left(\frac{3x}{2}\right)dx=ds$$$$I=\frac{2}{3}\int \frac{{(1+s)}^2}{\left(\frac{{(1+s)}^2}{1+s^2}\right)}\frac{ds}{(1+s^2)}$$$$I=\frac{2}{3}\int ds=\frac{2}{3}s+c$$$$I=\frac{2}{3}\tan \left(\frac{3x}{2}\right)+c$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com