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

From Wikibooks, open books for an open world
Jump to: navigation, search
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}