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 147837 by peter frank last updated on 23/Jul/21

$$Evaluate$$$$\int \frac{\sin {}^8\theta -\cos {}^8\theta }{1-2\sin {}^2\theta \cos {}^2\theta }d\theta$$$$$$

Answered by liberty last updated on 24/Jul/21

$$\sin {}^{8}x-\cos {}^{8}x=(\sin {}^{4}x+\cos {}^{4}x)(\sin {}^{4}x-\cos {}^{4}x)$$$$=(1-2\sin {}^{2}x\cos {}^{2}x)(\sin {}^{2}x-\cos {}^{2}x)$$$$I=\int \frac{(1-2\sin {}^{2}x\cos {}^{2}x)(\sin {}^{2}x-\cos {}^{2}x)}{1-2\sin {}^{2}x\cos {}^{2}x}dx$$$$=\int -\cos 2xdx$$$$=-\frac{1}{2}\sin 2x+c$$$$=-\sin x\cos x+c$$

Commented by peter frank last updated on 24/Jul/21

$$thankyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com