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 157257 by john_santu last updated on 21/Oct/21

$$\int \frac{\sin {}^{6}x+\cos {}^{5}x}{\sin {}^{2}x\cos {}^{2}x}dx$$

Answered by qaz last updated on 21/Oct/21

$$\int \frac{\sin {}^6\mathrm{x}+\cos {}^5\mathrm{x}}{\sin {}^2\mathrm{xcos}{}^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{\sin {}^4\mathrm{x}}{\cos {}^2\mathrm{x}}\mathrm{dx}+\int \frac{\cos {}^3\mathrm{x}}{\sin {}^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{1-2\cos {}^2\mathrm{x}+\cos {}^4\mathrm{x}}{\cos {}^2\mathrm{x}}\mathrm{dx}+\int \frac{1-\sin {}^2\mathrm{x}}{\sin {}^2\mathrm{x}}\mathrm{d}\left(\sin \mathrm{x}\right)$$$$=\tan \mathrm{x}-2\mathrm{x}+\frac{1}{2}\int (1+\cos 2\mathrm{x})\mathrm{dx}-\frac{1}{\sin \mathrm{x}}-\sin \mathrm{x}$$$$=\tan \mathrm{x}-\frac{3}{2}\mathrm{x}+\frac{1}{4}\sin 2\mathrm{x}-\frac{1}{\sin \mathrm{x}}-\sin \mathrm{x}+\mathrm{C}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com