chezmarylou1i

2021-12-23

Show the steps needed to find the given solutions.
$\int \frac{\mathrm{arctan}c}{1+{x}^{2}}dx$

Travis Hicks

Expert

1) $\int \frac{\mathrm{arctan}\left(x\right)}{1+{x}^{2}}dx$
Let $u=\mathrm{arctan}\left(x\right)$
$\frac{du}{dx}=\frac{d}{dx}\left(\mathrm{arctan}\left(x\right)\right)$
$\frac{du}{dx}=\frac{1}{1+{x}^{2}}⇒dx=du\left(1+{x}^{2}\right)$
Substiute dx value in eq. 1
$⇒\int \frac{u\cdot du\left(1+{x}^{2}\right)}{\left(1+{x}^{2}\right)}=\int udu=\frac{{u}^{2}}{2}+C$
Substitute "u" $⇒\frac{{\left(\mathrm{arctan}\left(x\right)\right)}^{2}}{2}+C$

Jeffery Autrey

Expert

2) $\int {\mathrm{cos}}^{2}x\mathrm{sin}xdx$
Let $u=\mathrm{cos}x⇒\frac{du}{dx}=\frac{d}{dx}\left(\mathrm{cos}x\right)$
$=-\mathrm{sin}x$
Substitute dx "value inequality" $\frac{du}{-\mathrm{sin}x}=dx$
$\int {u}^{3}\mathrm{sin}x\frac{du}{-\mathrm{sin}x}=\int -{u}^{2}du=\frac{-{u}^{4}}{4}+C$
Substitute "u" value
$=\frac{-{\mathrm{cos}}^{4}x}{4}+C$

karton

Expert

3) $\int {\mathrm{sec}}^{2}x{\mathrm{tan}}^{3}xdx$
Let $u=\mathrm{tan}x:⇒\frac{du}{dx}=\frac{d}{dx}\left(\mathrm{tan}x\right)$
$dx=\frac{1}{{\mathrm{sec}}^{2}x}du$
Substitute "dx" value in equation
$\int {\mathrm{sec}}^{2}x{u}^{2}\frac{du}{{\mathrm{sec}}^{2}x}=\frac{{u}^{4}}{4}+C=\frac{{\mathrm{tan}}^{4}\left(x\right)}{4}+C$

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