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 119934 by bramlexs22 last updated on 28/Oct/20

$$\int \frac{\sin {}^{8}x-\cos {}^{8}x}{1-\frac{1}{2}\sin {}^{2}2x}dx$$

Answered by bobhans last updated on 28/Oct/20

$$\int \frac{(\sin {}^{4}x-\cos {}^{4}x)(\sin {}^{4}x+\cos {}^{4}x)}{1-2\sin {}^{2}x\cos {}^{2}x}dx=$$$$\int \frac{(1-2\sin {}^{2}x\cos {}^{2}x)\left(1\right)(\sin {}^{2}x-\cos {}^{2}x)}{1-2\sin {}^{2}x\cos {}^{2}x}dx\stackrel{}{}$$$$\int -\cos 2xdx=-\frac{1}{2}\sin 2x+C$$$$=-\sin x\cos x+C$$

Answered by mathmax by abdo last updated on 28/Oct/20

$$\mathrm{A}=\int \frac{{\sin }^8\mathrm{x}-{\cos }^8\mathrm{x}}{1-\frac{1}{2}{\sin }^2\left(2\mathrm{x}\right)}\mathrm{dx}\Rightarrow \mathrm{A}=\int \frac{{\left({\sin }^4\mathrm{x}\right)}^2-{\left({\cos }^4\mathrm{x}\right)}^2}{1-\frac{1}{2}{\sin }^2\left(2\mathrm{x}\right)}\mathrm{dx}$$$$=\int \frac{({\sin }^4\mathrm{x}-{\cos }^4\mathrm{x})({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x})}{1-\frac{1}{2}{\sin }^2\left(2\mathrm{x}\right)}\mathrm{dx}$$$$=\int \frac{({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x})\left(1\right)(1-{2\sin }^2\mathrm{x}{\cos }^2\mathrm{x})}{1-\frac{1}{2}{\sin }^2\left(2\mathrm{x}\right)}\mathrm{dx}$$$$=\int \frac{({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x})(1-\frac{1}{2}{\sin }^2\left(2\mathrm{x}\right))}{1-\frac{1}{2}{\sin }^2\left(2\mathrm{x}\right)}\mathrm{dx}=-\int \cos \left(2\mathrm{x}\right)$$$$=-\frac{1}{2}\sin \left(2\mathrm{x}\right)+\mathrm{C}=-\mathrm{sinx}\mathrm{cosx}+\mathrm{C}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com