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 111029 by Khanacademy last updated on 01/Sep/20

Answered by mindispower last updated on 02/Sep/20

$$={\int }_0^{\pi }\frac{{(\pi -x)}^{2}sin(2\pi -2x)sin\left(\frac{\pi }{2}cos(\pi -x)\right)}{\pi -2x}dx$$$$=-{\int }_0^{\pi }\frac{({\pi }^2-2x\pi +x^2)sin\left(2x\right)sin\left(\frac{\pi }{2}cos\left(x\right)\right)}{2x-\pi }dx$$$$2{\int }_0^{\pi }\frac{x^{2}sin\left(2x\right)sin\left(\frac{\pi }{2}cos\left(x\right)\right)dx}{2x-\pi }=$$$$={\int }_0^{\pi }\frac{\pi (2x-\pi )sin\left(2x\right)sin\left(\frac{\pi }{2}cos\left(x\right)\right)dx}{2x-\pi }$$$$={\int }_0^{\pi }\pi sin\left(2x\right)sin\left(\frac{\pi }{2}cos\left(x\right)\right)dx$$$$=2\pi {\int }_0^{\pi }sin\left(x\right)cos\left(x\right)sin\left(\frac{\pi }{2}cos\left(x\right)\right)dx$$$$let\frac{\pi }{2}cos\left(x\right)=u$$$$\Rightarrow =\frac{8}{\pi }{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}usin\left(u\right)du=\frac{16}{\pi }{\int }_0^{\frac{\pi }{2}}usin\left(u\right)du$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com