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 26357 by abdo imad last updated on 24/Dec/17

$$findthevalueof{\int }_0^{\frac{\pi }{2}}ln\left(cos\theta \right)d\theta and$$$${\int }_0^{\frac{\pi }{2}}ln\left(sin\theta \right)d\theta .$$

Commented by prakash jain last updated on 24/Dec/17

$${\int }_0^{\pi /2}\ln \left(\cos x\right)dx$$$$u=\frac{\pi }{2}-x$$$$du=-dx$$$$I={\int }_0^{\pi /2}\ln \left(\sin x\right)dx=-{\int }_{\pi /2}^0\ln \left(\cos u\right)du$$$$={\int }_0^{\pi /2}\ln \left(\cos x\right)dx$$$$2I={\int }_0^{\pi /2}\ln \left(\cos x\sin x\right)dx$$$$={\int }_0^{\pi /2}\ln \sin 2xdx-{\int }_0^{\pi /2}\ln 2dx$$$$=I-\frac{\pi }{2}\ln 2$$$$I=-\frac{\pi }{2}\ln 2$$

Answered by prakash jain last updated on 24/Dec/17

$$-\frac{\pi }{2}\ln 2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com