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

1. $\int\frac{19}{\sqrt{9x-38}}dx$

Let

$u=9x-38\qquad du=9dx\qquad dx=\frac{du}{9}$

Then

\begin{align}\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{align}
2. $\int-15\sqrt{9x+43}dx$

Let

$u=9x+43\qquad du=9dx\qquad dx=\frac{du}{9}$

Then

\begin{align}\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{align}
3. $\int\frac{17\sin(x)}{\cos(x)}dx$

Let

$u=\cos(x)\qquad du=-\sin(x)dx\qquad dx=-\frac{du}{\sin(x)}$

Then

\begin{align}\int\frac{17\sin(x)}{\cos(x)}dx&=17\int-\frac{du}{u}\\ &=-17\ln|u|+C\\ &=\mathbf{-17\ln|\cos(x)|+C}\end{align}
4. $\int5\cos(x)\sin(x)dx$

Let

$u=\sin(x)\qquad du=\cos(x)dx\qquad dx=\frac{du}{\cos(x)}$

Then

\begin{align}\int5\cos(x)\sin(x)dx&=5\int udu\\ &=5\frac{u^{2}}{2}+C\\ &=\mathbf{\frac{5\sin^{2}(x)}{2}+C}\end{align}
5. $\int_{0}^{1}-\frac{10}{(-5x-32)^{4}}dx$

Let

$u=-5x-32\qquad du=-5dx\qquad dx=-\frac{du}{5}$

Then

\begin{align}\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}\cdot37^{3}}\\ &=\frac{2}{3}\cdot\frac{32^{3}-37^{3}}{2^{15}\cdot37^{3}}\\ &=\frac{32^{3}-37^{3}}{2^{14}\cdot3\cdot37^{3}}\\ &=\mathbf{-\frac{17885}{2489696256}}\end{align}
6. $\int-3e^{3x+12}dx$

Let

$u=3x+12\qquad du=3dx\qquad dx=\frac{du}{3}$

Then

\begin{align}\int-3e^{3x+12}dx&=-3\int\frac{e^{u}}{3}du\\ &=-e^{u}+C\\ &=\mathbf{-e^{3x+12}+C}\end{align}
Last modified on 27 September 2011, at 20:10