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 185284 by Mingma last updated on 19/Jan/23

Answered by MJS_new last updated on 20/Jan/23

$$\mathrm{one}\mathrm{method}:$$$$\int \frac{\sin x\cos x}{{\sin }^{3}x+{\cos }^{3}x}dx=$$$$[t=\cos (x+\frac{\pi }{4})\to dx=-\frac{dt}{\sin (x+\frac{\pi }{4})}]$$$$=-\frac{\sqrt{2}}{2}\int \frac{2t^2-1}{(t-1)(t+1)(2t^2+1)}dt=$$$$=\frac{\sqrt{2}}{12}\int \frac{dt}{t+1}-\frac{\sqrt{2}}{12}\int \frac{dt}{t-1}-\frac{2\sqrt{2}}{3}\int \frac{dt}{2t^2+1}=$$$$=\frac{\sqrt{2}}{12}\ln (t+1)-\frac{\sqrt{2}}{12}\ln (t-1)-\frac{2}{3}\arctan \sqrt{2}t=$$$$=\frac{\sqrt{2}}{12}\ln \mid \frac{\sin x-\cos x-\sqrt{2}}{\sin x+\cos x+\sqrt{2}}\mid +\frac{2}{3}\arctan (\sin x-\cos x)+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com