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 149113 by bramlexs22 last updated on 03/Aug/21

$${\int }_0^{\pi }\frac{\cos {}^3\mathrm{x}}{7-\sin {}^2\mathrm{x}}\mathrm{dx}=?$$

Answered by mathmax by abdo last updated on 03/Aug/21

$$\mathrm{\Psi }={\int }_0^{\pi }\frac{{\cos }^3\mathrm{x}}{7-{\sin }^2\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{\Psi }={\int }_0^{\pi }\frac{{\cos }^3\mathrm{x}}{7-(1-{\cos }^2\mathrm{x})}\mathrm{dx}$$$$={\int }_0^{\pi }\frac{{\cos }^3\mathrm{x}}{6+{\cos }^2\mathrm{x}}\mathrm{dx}={\int }_0^{\pi }\frac{\mathrm{cosx}({\cos }^2\mathrm{x}+6)-6\mathrm{cosx}}{{\cos }^2\mathrm{x}+6}\mathrm{dx}$$$$={\int }_0^{\pi }\mathrm{cosx}\mathrm{dx}-6{\int }_0^{\pi }\frac{\mathrm{cosx}}{7-{\sin }^2\mathrm{x}}\mathrm{dx}\mathrm{we}\mathrm{have}$$$${\int }_0^{\pi }\mathrm{cosx}\mathrm{dx}={\left[\mathrm{sinx}\right]}_0^{\pi }=0\mathrm{and}$$$${\int }_0^{\pi }\frac{\mathrm{cosxdx}}{7-{\sin }^2\mathrm{x}}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cosx}}{7-{\sin }^2\mathrm{x}}\mathrm{dx}+{\int }_{\frac{\pi }{2}}^{\pi }\frac{\mathrm{cosx}}{7-{\sin }^2\mathrm{x}}(\to \mathrm{x}=\frac{\pi }{2}+\mathrm{t})$$$${=}_{\mathrm{sinx}=\mathrm{y}}{\int }_0^1\frac{\mathrm{dy}}{7-{\mathrm{y}}^2}+{\int }_0^{\frac{\pi }{2}}\frac{-\mathrm{sint}\mathrm{dt}}{7-{\cos }^2\mathrm{t}}(\to \mathrm{cost}=\mathrm{y})$$$$={\int }_0^1\frac{\mathrm{dy}}{7-{\mathrm{y}}^2}-{\int }_0^1\frac{\mathrm{dy}}{7-{\mathrm{y}}^2}=0\Rightarrow \mathrm{\Psi }=0$$

Commented by bramlexs22 last updated on 03/Aug/21

$$\mathrm{yes}.\mathrm{thanks}$$

Commented by ArielVyny last updated on 04/Aug/21

$$mrmathmaxwhenwetaket=tan\left(x\right)$$$$whatisthevalueofcosxandsinx$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com