# Test Colors

 $$\boldsymbol {U}$$-Sub Integral Setup Solution and Check $$\displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx$$ $$\displaystyle \begin{array}{c}\color{blue}{{u=2{{x}^{2}}+3}}\\du=4x\,dx\end{array}$$ $$\displaystyle dx=\frac{{du}}{{4x}}$$ $$\require {cancel} \displaystyle \int{{{{{\left( {\color{blue}{{2{{x}^{2}}+3}}} \right)}}^{5}}}}x\,\color{green}{{dx}}=\int{{{{{\color{blue}{u}}}^{5}}}}\cancel{x}\color{green}{{\frac{{du}}{{4\cancel{x}}}}}=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{5}}\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{6}{{\color{blue}{u}}^{6}}+C=\frac{1}{{24}}{{\color{blue}{u}}^{6}}+C$$ $$\displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx=\frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{6}}+C=\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}+C$$   Differentiate back to verify: \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}} \right)&=6\cdot \frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot 4x\\&=\frac{1}{{\cancel{4}}}{{\left( {2{{x}^{2}}+3} \right)}^{5}}\cdot \cancel{4}x={{\left( {2{{x}^{2}}+3} \right)}^{5}}x\end{align}    √ $$\int{{\cos \left( {4x} \right)\,{{{\sin }}^{3}}}}\left( {4x} \right)\,dx$$     Note that this is actually: $$\displaystyle \int{{\cos \left( {4x} \right)\,}}{{\left( {\sin \left( {4x} \right)} \right)}^{3}}dx$$ \displaystyle \begin{align}&\color{blue}{{u=\sin \left( {4x} \right)}}\,\,\\du&=4\cos \left( {4x} \right)\,dx\\&\color{green}{{dx=\frac{{du}}{{4\cos \left( {4x} \right)}}}}\end{align} \displaystyle \begin{align}\int{{\cos \left( {4x} \right){{{\color{blue}{{\sin }}}}^{3}}\left( {\color{blue}{{4x}}} \right)}}\,\color{green}{{dx}}&=\int{{\cancel{{\cos \left( {4x} \right)}}{{{\color{blue}{u}}}^{3}}}}\,\color{green}{{\frac{{du}}{{4\cancel{{\cos \left( {4x} \right)}}}}}}\\&=\frac{1}{4}\int{{{{{\color{blue}{u}}}^{3}}\,\color{green}{{du}}}}=\frac{1}{4}\cdot \frac{1}{4}{{\color{blue}{u}}^{4}}+C=\frac{1}{{16}}{{\color{blue}{u}}^{4}}+C\end{align} $$\displaystyle \int{{\cos \left( {4x} \right){{{\sin }}^{3}}\left( {4x} \right)}}\,dx=\frac{1}{{16}}{{\sin }^{4}}\left( {4x} \right)+C$$   Differentiate back to verify: \displaystyle \begin{align}\frac{d}{{dx}}\left( {\frac{1}{{16}}{{{\sin }}^{4}}\left( {4x} \right)} \right)&=\frac{1}{{16}}\cdot 4\sin {{\left( {4x} \right)}^{3}}\cdot \cos \left( {4x} \right)\cdot 4\\&=\cancel{{\frac{{16}}{{16}}}}\sin {{\left( {4x} \right)}^{3}}\cos \left( {4x} \right)=\cos \left( {4x} \right)\sin {{\left( {4x} \right)}^{3}}\end{align}       √ $$\displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}$$ \begin{align}\color{blue}{{u=3{{x}^{2}}+2}}\\du=6x\,dx\\\color{green}{{dx=\frac{{du}}{{6x}}}}\end{align} \displaystyle \begin{align} \int{{\frac{{x\,\color{green}{dx}}}{{\sqrt{{\color{blue}{{3{{x}^{2}}+2}}}}}}}}&=\int{{{{{\left( {\color{blue}{{3{{x}^{2}}+2}}} \right)}}^{{-\frac{1}{2}}}}x\,\color{green}{{dx}}=}}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}}}\cancel{x}\color{green}{{\frac{{du}}{{6\cancel{x}}}}}=\frac{1}{6}\int{{{{{\color{blue}{u}}}^{{-\frac{1}{2}}}}\color{green}{du}}}\\&=\frac{{\left( {\frac{1}{6}} \right)}}{{\left( {\frac{1}{2}} \right)}}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C=-\frac{1}{3}{{\color{blue}{u}}^{{\frac{1}{2}}}}+C\end{align} $$\displaystyle \int{{\frac{{x\,\,dx}}{{\sqrt{{3{{x}^{2}}+2}}}}}}x\,dx=\frac{1}{3}{{\left( {3{{x}^{2}}+2} \right)}^{{\frac{1}{2}}}}+C=\frac{{\sqrt{{3{{x}^{2}}+2}}}}{3}\,+\,C$$   Differentiate back to verify: $$\displaystyle \frac{{d\left( {\frac{{\sqrt{{3{{x}^{2}}+2}}}}{3}} \right)}}{{dx}}\,=\frac{{d\left( {\frac{1}{3}{{{\left( {3{{x}^{2}}+2} \right)}}^{{\frac{1}{2}}}}} \right)}}{{dx}}=\frac{1}{{\cancel{6}}}{{\left( {3{{x}^{2}}+2} \right)}^{{-\frac{1}{2}}}}\cdot \cancel{6}x=\frac{x}{{\sqrt{{3{{x}^{2}}+2}}}}\,$$      √

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 $$\boldsymbol {U}$$-Sub Integral $$u=g\left( x \right)$$ $$\displaystyle du={g}’\left( x \right)dx$$ Substitute, Integrate, and Substitute Back! $$\displaystyle \int{{{{{\left( {\color{blue}{{5{{x}^{2}}+3}}} \right)}}^{4}}}}\color{red}{{\left( {10x} \right)dx}}$$ $$u=5{{x}^{2}}+3$$ $$\displaystyle du=\left( {10x} \right)dx$$ $$\displaystyle \int{{{{{\color{blue}{{\left( {5{{x}^{2}}+3} \right)}}}}^{4}}}}\color{red}{{\left( {10x} \right)dx}}=\int{{{{{\color{blue}{u}}}^{4}}}}\,\color{red}{{du}}=\frac{1}{5}{{u}^{5}}+C=\frac{1}{5}{{\left( {5{{x}^{2}}+3} \right)}^{5}}+C$$ $$\displaystyle \int{{\color{red}{{{{{\sec }}^{2}}x}}\sqrt{{\color{blue}{{\tan x}}}}}}\,\color{red}{{dx}}$$ $$u=\tan x$$ $$\displaystyle \int{{{{{\left( {5{{x}^{2}}+3} \right)}}^{4}}}}\left( {10x} \right)dx$$ $$\displaystyle \int{{\color{red}{{{{{\sec }}^{2}}x}}\sqrt{{\color{blue}{{\tan x}}}}}}\,\color{red}{{dx}}=\int{{\sqrt{{\color{blue}{u}}}\,}}\color{red}{{du}}=\int{{{{{\color{blue}{u}}}^{{\frac{1}{2}}}}}}\,\color{red}{{du}}=\frac{2}{3}{{u}^{{\frac{3}{2}}}}+C=\frac{2}{3}{{\left( {\tan x} \right)}^{{\frac{3}{2}}}}+C$$ $$\displaystyle \int{{{{{\left( {2{{x}^{2}}+3} \right)}}^{5}}}}x\,dx=\frac{1}{{24}}{{\left( {2{{x}^{2}}+3} \right)}^{6}}+C=\frac{{{{{\left( {2{{x}^{2}}+3} \right)}}^{6}}}}{{24}}+C$$
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