# Evaluate the following1.int e^x sqrt (1-e^(2x)) dx2. int frac{x^2}{sqrt 5+x^2}dx

Evaluate the following
1.$\int {e}^{x}\sqrt{1-{e}^{2x}}dx$
2. $\int \frac{{x}^{2}}{\sqrt{5}+{x}^{2}}dx$
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pattererX
1.Finding the integral $\int {e}^{x}\sqrt{1-{e}^{2x}dx.}$
Let us first apply substitution $u={e}^{x}$ we get
$\int {e}^{x}\sqrt{1-{e}^{2x}}dx=\int \sqrt{1-{u}^{2}}du$
Further apply the substitution $u=\mathrm{sin}v$ we get
$\int \sqrt{1-{u}^{2}}du=\int {\mathrm{cos}}^{2}\left(v\right)dv$
=$\int \frac{1+\mathrm{cos}\left(2v\right)}{2}dv\left[:‘{\mathrm{cos}}^{2}\left(x\right)=\frac{1+\mathrm{cos}\left(2x\right)}{2}\right]$
= $\frac{1}{2}\cdot \left(v+\frac{1}{2}\mathrm{sin}\left(2v\right)\right)$
Substitute back $v=\mathrm{arcsin}\left(u\right)$, and $u={e}^{x}$, we get $v=\mathrm{arcsin}\left({e}^{x}\right)$,we get $\int {e}^{x}\sqrt{1-{e}^{2x}}dx=\frac{1}{2}\left(\mathrm{arcsin}\left({e}^{x}\right)+\frac{1}{2}\mathrm{sin}\left(2\mathrm{arcsin}\left({e}^{x}\right)\right)\right)+C$
2. Finding the integral $\int \frac{{x}^{2}}{\sqrt{5}+{x}^{2}}dx$
$\int \frac{{x}^{2}}{\sqrt{5}+{x}^{2}}dx=5\cdot \int {\mathrm{tan}}^{2}\left(u\right)\mathrm{sec}\left(u\right)du\left[x=\sqrt{5}\mathrm{tan}\left(u\right)\right]$
$=5\cdot \int \left(-1+{\mathrm{sec}}^{2}\left(u\right)\right)\mathrm{sec}\left(u\right)du\left[{\mathrm{tan}}^{2}\left(x\right)=-1+{\mathrm{sec}}^{2}\left(x\right)\right]$
=$5\left(\mathrm{ln}|\mathrm{tan}\left(u\right)+\mathrm{sec}\left(u\right)|+\frac{1}{2}\mathrm{sec}\left(u\right)\mathrm{tan}\left(u\right)+\frac{1}{2}\mathrm{ln}\mid \mathrm{tan}\left(u\right)+\mathrm{sec}\right)$
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