Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 122551 by shaker last updated on 18/Nov/20

Answered by bramlexs22 last updated on 18/Nov/20

$$\int \frac{4\sin {}^{2}2x.\cos {}^{2}2x}{2\cos {}^{2}x}dx=2\int \frac{4\sin {}^{2}x\cos {}^{2}x.\cos {}^{2}2x}{\cos {}^{2}x}dx$$$$=8\int \sin {}^{2}x\cos {}^{2}2xdx$$$$=8\int (\frac{1}{2}-\frac{1}{2}\cos 2x)(\frac{1}{2}+\frac{1}{2}\cos 4x)dx$$$$=2\int (1-\cos 2x)(1+\cos 4x)dx$$$$=2\int 1+\cos 4x-\cos 2x-\frac{1}{2}(\cos 6x+\cos 2x))dx$$$$=2(x+\frac{\sin 4x}{4}-\frac{3\sin 2x}{4}-\frac{\sin 6x}{12})+c$$

Commented by MJS_new last updated on 18/Nov/20

$$\mathrm{error}\mathrm{in}\mathrm{line}4$$$$\mathrm{it}\mathrm{must}\mathrm{be}2\int (1-\cos 2x)(1+\cos 4x)dx$$$$\mathrm{I}\mathrm{get}$$$$2x-\frac{\sin 6x}{6}+\frac{\sin 4x}{2}-\frac{3\sin 2x}{2}+C$$$$\mathrm{so}\mathrm{something}\mathrm{else}\mathrm{went}\mathrm{wrong}$$

Commented by bramlexs22 last updated on 18/Nov/20

$$ooyes.thankyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com