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 183156 by cortano1 last updated on 21/Dec/22

$$\int \frac{\sin 2xdx}{\sin x-\sin {}^{2}2x}=?$$

Answered by MJS_new last updated on 21/Dec/22

$$\int \frac{\sin 2x}{\sin x-{\sin }^{2}2x}dx=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=\frac{1}{2}\int \frac{dt}{t^3-t+\frac{1}{4}}=\frac{1}{2}\int \frac{dt}{(t-\alpha )(t-\beta )(t-\gamma )}$$$$\mathrm{now}\mathrm{use}\mathrm{the}\mathrm{usual}\mathrm{method}$$$$\mathrm{with}$$$$\alpha =-\frac{2\sqrt{3}}{3}\sin \frac{\pi +\arcsin \frac{3\sqrt{3}}{8}}{3}$$$$\beta =\frac{2\sqrt{3}}{3}\sin \frac{\arcsin \frac{3\sqrt{3}}{8}}{3}$$$$\gamma =\frac{2\sqrt{3}}{3}\cos \frac{\pi +2\arcsin \frac{3\sqrt{3}}{8}}{6}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com