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Question Number 68043 by mhmd last updated on 03/Sep/19

$${\int }_{\pi /2}^{\pi }{e}^{cosx}\sqrt{1-e^{cosx}}sinxdx$$

Commented by mathmax by abdo last updated on 03/Sep/19

$$letI={\int }_{\frac{\pi }{2}}^{\pi }{e}^{cosx}\sqrt{1-e^{cosx}}sinxdxvha7gementcosx=-tgive$$$$I={\int }_0^1{e}^{-t}\sqrt{1-e^{-t}}\left(dt\right)={\int }_0^1{e}^{-t}\sqrt{1-e^{-t}}dtafterwedothe$$$$changement\sqrt{1-e_{}^{-t}}=u\Rightarrow 1-e^{-t}=u^2\Rightarrow e^{-t}=1-u^2\Rightarrow$$$$-t=ln(1-u^2)\Rightarrow t=-ln(1-u^2)\Rightarrow$$$$I={\int }_0^{\sqrt{1-e^{-1}}}(1-u^2)u\times \frac{2u}{1-u^2}du=2{\int }_0^{\sqrt{1-e^{-1}}}{u}^{2}du$$$$=\frac{2}{3}{\left[u^3\right]}_0^{\sqrt{1-e^{-1}}}=\frac{2}{3}{\left\{\sqrt{1-e^{-1}}\right\}}^3=\frac{2}{3}(1-e^{-1})\sqrt{1-\frac{1}{e}}$$$$=\frac{2}{3}(1-\frac{1}{e})\frac{\sqrt{e-1}}{\sqrt{e}}=\frac{2}{3}\left(\frac{e-1}{e}\right)\frac{\sqrt{e-1}}{\sqrt{e}}$$$$\Rightarrow I=\frac{2(e-1)\sqrt{e-1}}{3e\sqrt{e}}.$$

Answered by mr W last updated on 03/Sep/19

$$=-{\int }_{\pi /2}^{\pi }{e}^{cosx}\sqrt{1-e^{cosx}}d\cos x$$$$={\int }_{-1}^0{e}^t\sqrt{1-e^t}dt$$$$={\int }_{-1}^0\sqrt{1-e^t}{de}^t$$$$={\int }_{\frac{1}{e}}^1\sqrt{1-u}du$$$$=-{\int }_{\frac{1}{e}}^1\sqrt{1-u}d(1-u)$$$$=-\frac{2}{3}{\left[{(1-u)}^{\frac{3}{2}}\right]}_{\frac{1}{e}}^1$$$$=\frac{2}{3}{(1-\frac{1}{e})}^{\frac{3}{2}}$$

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