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Question Number 35202 by Cheyboy last updated on 16/May/18

$${\int }^{}{\mathit{e}}^{3\mathit{x}}\sqrt{1-{\mathit{e}}^{2\mathit{x}}}\mathit{d}\mathit{x}$$$$\mathit{p}\mathit{l}\mathit{z}\mathit{z}\mathit{h}\mathit{e}\mathit{l}\mathit{p}$$

Answered by ajfour last updated on 16/May/18

$$let{e}^x=t\Rightarrow e^{x}dx=dt$$$$I=\int e^{2x}\sqrt{1-e^{2x}}{e}^{x}dx=\int t^2\sqrt{1-t^2}dt$$$$nowlett=\sin \theta \Rightarrow dt=\cos \theta d\theta$$$$I=\int \sin {}^2\theta \cos {}^2\theta d\theta$$$$=\frac{1}{8}\int (1-\cos 4\theta )d\theta$$$$=\frac{\theta }{8}-\frac{\sin 4\theta }{32}+c$$$$=\frac{{\sin }^{-1}\left(e^x\right)}{8}-\frac{1}{32}\sin \left[{4\sin }^{-1}\left(e^x\right)\right]+c.$$

Commented by Cheyboy last updated on 16/May/18

$$\mathcal{T}hankxalotsir$$

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