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 169855 by Shrinava last updated on 11/May/22

Answered by thfchristopher last updated on 11/May/22

$$={\int }_0^{\frac{\pi }{4}}\sin (3x-x)dx$$$$={\int }_0^{\frac{\pi }{4}}\sin 2xdx$$$$-\frac{1}{2}{\left[\cos 2x\right]}_0^{\frac{\pi }{4}}$$$$\frac{1}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com