# Calculus/Integration techniques/Recognizing Derivatives and the Substitution Rule/Solutions

$\int {\frac {19}{\sqrt {9x-38}}}dx$
    $u=9x-38\qquad du=9dx\qquad dx={\frac {du}{9}}$
Then
    ${\begin{aligned}\int {\frac {19}{\sqrt {9x-38}}}dx&=\int {\frac {19}{9{\sqrt {u}}}}du\\&={\frac {19}{9}}(2{\sqrt {u}})+C\\&=\mathbf {{\frac {38{\sqrt {9x-38}}}{9}}+C} \end{aligned}}$
    $u=9x-38\qquad du=9dx\qquad dx={\frac {du}{9}}$
Then
    ${\begin{aligned}\int {\frac {19}{\sqrt {9x-38}}}dx&=\int {\frac {19}{9{\sqrt {u}}}}du\\&={\frac {19}{9}}(2{\sqrt {u}})+C\\&=\mathbf {{\frac {38{\sqrt {9x-38}}}{9}}+C} \end{aligned}}$
$\int -15{\sqrt {9x+43}}dx$
    $u=9x+43\qquad du=9dx\qquad dx={\frac {du}{9}}$
Then
    ${\begin{aligned}\int -15{\sqrt {9x+43}}dx&=-15\int {\frac {\sqrt {u}}{9}}du\\&=-{\frac {15}{9}}{\frac {2}{3}}u^{3/2}+C\\&=\mathbf {-{\frac {10(9x+43)^{3/2}}{9}}+C} \end{aligned}}$
    $u=9x+43\qquad du=9dx\qquad dx={\frac {du}{9}}$
Then
    ${\begin{aligned}\int -15{\sqrt {9x+43}}dx&=-15\int {\frac {\sqrt {u}}{9}}du\\&=-{\frac {15}{9}}{\frac {2}{3}}u^{3/2}+C\\&=\mathbf {-{\frac {10(9x+43)^{3/2}}{9}}+C} \end{aligned}}$
$\int {\frac {17\sin(x)}{\cos(x)}}dx$
    $u=\cos(x)\qquad du=-\sin(x)dx\qquad dx=-{\frac {du}{\sin(x)}}$
Then
    ${\begin{aligned}\int {\frac {17\sin(x)}{\cos(x)}}dx&=17\int -{\frac {du}{u}}\\&=-17\ln |u|+C\\&=\mathbf {-17\ln |\cos(x)|+C} \end{aligned}}$
    $u=\cos(x)\qquad du=-\sin(x)dx\qquad dx=-{\frac {du}{\sin(x)}}$
Then
    ${\begin{aligned}\int {\frac {17\sin(x)}{\cos(x)}}dx&=17\int -{\frac {du}{u}}\\&=-17\ln |u|+C\\&=\mathbf {-17\ln |\cos(x)|+C} \end{aligned}}$
$\int 5\cos(x)\sin(x)dx$
    $u=\sin(x)\qquad du=\cos(x)dx\qquad dx={\frac {du}{\cos(x)}}$
Then
    ${\begin{aligned}\int 5\cos(x)\sin(x)dx&=5\int udu\\&=5{\frac {u^{2}}{2}}+C\\&=\mathbf {{\frac {5\sin ^{2}(x)}{2}}+C} \end{aligned}}$
    $u=\sin(x)\qquad du=\cos(x)dx\qquad dx={\frac {du}{\cos(x)}}$
Then
    ${\begin{aligned}\int 5\cos(x)\sin(x)dx&=5\int udu\\&=5{\frac {u^{2}}{2}}+C\\&=\mathbf {{\frac {5\sin ^{2}(x)}{2}}+C} \end{aligned}}$
$\int _{0}^{1}-{\frac {10}{(-5x-32)^{4}}}dx$
    $u=-5x-32\qquad du=-5dx\qquad dx=-{\frac {du}{5}}$
Then
    ${\begin{aligned}\int _{0}^{1}-{\frac {10}{(-5x-32)^{4}}}dx&=-10\int _{u(0)}^{u(1)}{\frac {-du}{5u^{4}}}\\&=-2{\frac {1}{3u^{3}}}{\Biggr |}_{u(0)}^{u(1)}\\&=-2{\frac {1}{3(-5x-32)^{3}}}{\Biggr |}_{0}^{1}\\&=-{\frac {2}{3}}({\frac {1}{(-5-32)^{3}}}-{\frac {1}{(-32)^{3}}})\\&=-{\frac {2}{3}}({\frac {1}{(-37)^{3}}}+{\frac {1}{(32)^{3}}})\\&={\frac {2}{3}}({\frac {1}{37^{3}}}-{\frac {1}{32^{3}}})\\&={\frac {2}{3}}\cdot {\frac {32^{3}-37^{3}}{32^{3}\cdot 37^{3}}}\\&={\frac {2}{3}}\cdot {\frac {32^{3}-37^{3}}{2^{15}\cdot 37^{3}}}\\&={\frac {32^{3}-37^{3}}{2^{14}\cdot 3\cdot 37^{3}}}\\&=\mathbf {-{\frac {17885}{2489696256}}} \end{aligned}}$
    $u=-5x-32\qquad du=-5dx\qquad dx=-{\frac {du}{5}}$
Then
    ${\begin{aligned}\int _{0}^{1}-{\frac {10}{(-5x-32)^{4}}}dx&=-10\int _{u(0)}^{u(1)}{\frac {-du}{5u^{4}}}\\&=-2{\frac {1}{3u^{3}}}{\Biggr |}_{u(0)}^{u(1)}\\&=-2{\frac {1}{3(-5x-32)^{3}}}{\Biggr |}_{0}^{1}\\&=-{\frac {2}{3}}({\frac {1}{(-5-32)^{3}}}-{\frac {1}{(-32)^{3}}})\\&=-{\frac {2}{3}}({\frac {1}{(-37)^{3}}}+{\frac {1}{(32)^{3}}})\\&={\frac {2}{3}}({\frac {1}{37^{3}}}-{\frac {1}{32^{3}}})\\&={\frac {2}{3}}\cdot {\frac {32^{3}-37^{3}}{32^{3}\cdot 37^{3}}}\\&={\frac {2}{3}}\cdot {\frac {32^{3}-37^{3}}{2^{15}\cdot 37^{3}}}\\&={\frac {32^{3}-37^{3}}{2^{14}\cdot 3\cdot 37^{3}}}\\&=\mathbf {-{\frac {17885}{2489696256}}} \end{aligned}}$
$\int -3e^{3x+12}dx$
    $u=3x+12\qquad du=3dx\qquad dx={\frac {du}{3}}$
Then
    ${\begin{aligned}\int -3e^{3x+12}dx&=-3\int {\frac {e^{u}}{3}}du\\&=-e^{u}+C\\&=\mathbf {-e^{3x+12}+C} \end{aligned}}$
    $u=3x+12\qquad du=3dx\qquad dx={\frac {du}{3}}$
Then
    ${\begin{aligned}\int -3e^{3x+12}dx&=-3\int {\frac {e^{u}}{3}}du\\&=-e^{u}+C\\&=\mathbf {-e^{3x+12}+C} \end{aligned}}$
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