Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 137784 by mathocean1 last updated on 06/Apr/21

$${\int }_0^{\frac{\pi }{6}}\frac{dx}{sinx}=?$$$$$$

Answered by EnterUsername last updated on 06/Apr/21

$$\int \frac{dx}{sinx}=ln\mid cosecx-cotx\mid +C$$$$\int \frac{dx}{sinx}=\int \frac{sinx}{{sin}^{2}x}dx=\int \frac{sinx}{1-{cos}^{2}x}dx=\int \frac{du}{u^2-1}$$$$=\int (\frac{1}{2(u-1)}-\frac{1}{2(u+1)})du=\frac{1}{2}ln\mid \frac{u-1}{u+1}\mid +C$$$$=\frac{1}{2}ln\mid \frac{cosx-1}{cosx+1}\mid +C=\frac{1}{2}ln\mid \frac{{(cosx-1)}^2}{{cos}^{2}x-1}\mid +C$$$$=\frac{1}{2}ln\mid \frac{{(cosx-1)}^2}{{sin}^{2}x}\mid +C=ln\mid \frac{cosx-1}{sinx}\mid +C$$$$=ln\mid cosecx-cotx\mid +C$$$${\int }_0^{\frac{\pi }{6}}\frac{dx}{sinx}=ln\mid 2-\sqrt{3}\mid -\underset{x\to 0}{lim}(ln\mid cosecx-cotx\mid )$$

Answered by mathmax by abdo last updated on 06/Apr/21

$${\int }_0^{\frac{\pi }{6}}\frac{\mathrm{dx}}{\mathrm{sinx}}={\lim }_{\xi \to 0}{\int }_{\xi }^{\frac{\pi }{6}}\frac{\mathrm{dx}}{\mathrm{sinx}}\mathrm{we}\mathrm{have}$$$${\int }_{\xi }^{\frac{\pi }{6}}\frac{\mathrm{dx}}{\mathrm{sinx}}{=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}{\int }_{\tan \left(\frac{\xi }{2}\right)}^{2-\sqrt{3}}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)\times \frac{2\mathrm{t}}{1+{\mathrm{t}}^2}}$$$$={\int }_{\tan \left(\frac{\xi }{2}\right)}^{2-\sqrt{3}}\frac{\mathrm{dt}}{\mathrm{t}}={[\log \mid \mathrm{t}\mid ]}_{\tan \left(\frac{\xi }{2}\right)}^{2-\sqrt{3}}=\log (2-\sqrt{3})-\log \mid \tan \left(\frac{\mathrm{x}}{2}\right)\mid \Rightarrow$$$${\lim }_{\xi \to 0}{\int }_{\xi }^{\frac{\pi }{6}}\frac{\mathrm{dx}}{\mathrm{sinx}}=+\mathrm{\infty }\mathrm{this}\mathrm{integral}\mathrm{is}\mathrm{divergent}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com