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Question Number 54856 by Meritguide1234 last updated on 13/Feb/19

Commented by Meritguide1234 last updated on 14/Feb/19

Commented by Meritguide1234 last updated on 14/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Feb/19

$$\frac{cosx}{1+2cosx}$$$$\frac{\frac{1}{2}cosx}{\frac{1}{2}+cosx}\to \frac{cos\frac{\pi }{3}cosx}{cos\frac{\pi }{3}+cosx}\to \frac{ab}{a+b}\to \frac{ab}{{(\sqrt{a}-\sqrt{b})}^2+2\sqrt{a}\sqrt{b}}$$$$maxvalueof\frac{ab}{a+b}when{D}_r=min\to \sqrt{a}-\sqrt{b}=0$$$$cosx=cos\frac{\pi }{3}\to x=\frac{\pi }{3}$$$${cos}^{-1}\left(\frac{cosx}{1+2cosx}\right)\to {cos}^{-1}\left(\frac{\frac{1}{2}}{1+2\times \frac{1}{2}}\right)\to {cos}^{-1}\left(\frac{1}{4}\right)$$$${cos}^{-1}\left(\frac{cosx}{1+2cosx}\right)\to {cos}^{-1}\left(\frac{1}{1+2\times 1}\right)atx=0$$$${cos}^{-1}\left(\frac{cos\frac{\pi }{2}}{1+2cos\frac{\pi }{2}}\right)\to {cos}^{-1}\left(0\right)atx=\frac{\pi }{2}$$$${cos}^{-}\left(0\right){\int }_0^{\frac{\pi }{2}}dx>{cos}^{-1}\left(\frac{1}{4}\right){\int }_0^{\frac{\pi }{2}}dx>{cos}^{-1}\left(\frac{1}{3}\right){\int }_0^{\frac{\pi }{2}}dx$$$$so\frac{\pi }{2}\times \frac{\pi }{2}>I>{cos}^{-1}\left(\frac{1}{3}\right)\times \frac{\pi }{2}$$$$plscheck...$$

Commented by Meritguide1234 last updated on 13/Feb/19

$$\mathrm{no}..\mathrm{ans}\mathrm{is}5{\pi }^2/24$$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

$$uploadtheanswerplswithworking...$$$$itriedtoreachuptothe....$$$$\frac{{\pi }^2}{4}>I>\frac{\pi }{2}{cos}^{-1}\left(\frac{1}{3}\right)$$$$\frac{6{\pi }^2}{24}>I>\frac{\pi }{2}{cos}^{-1}\left(\frac{1}{3}\right)$$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

$$thsnkyousir...letmeunlockthesecret...$$

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