Finding Indefinite Integrals Using Substitution

Find:

1. \(\int \sin (3x+2)\,dx \)
2. \(\int x\sqrt{x^{2}+1}\,dx\)
3. \(\int \dfrac{e^{\sqrt{{ x}}}}{\sqrt{x}}dx\)

Solution (a) Since we know \(\int \sin x\,dx,\) we let \(u = 3x + 2\). Then \(du = 3\,dx\) so \(dx = \dfrac{du}{3}\). \[ \begin{eqnarray*} \int \sin (\underset{\color{#0066A7}{\hbox{\(u\)}}}{\underbrace{(3x+2)}}\enspace\underset{\color{#0066A7}{\tfrac{du}{2}}}{\underbrace{dx}}&=& \int {\sin }\,u\dfrac{du}{3}=\dfrac{1}{3}\int {\sin }\,u\,du\\ &=&\dfrac{1}{3}({-}\cos u)+C \underset{\underset{{\color{#0066A7}{\hbox{\(u=3 x+2\)}}}}{\color{#0066A7}{\uparrow }}}{=} -\dfrac{1}{3}{\cos }(3x+2)+C \end{eqnarray*} \]

(b) We let \(u = x^{2} + 1\). Then \(du = 2x\,dx\), so \(x\,dx = \dfrac{du}{2}\). \[ \begin{eqnarray*} \int x\sqrt{x^{2}+1}\,dx &=& \int \underset{\color{#0066A7}{\hbox{\(u\)}}}{{\underbrace{\sqrt{x^{2}+1}}}\,}\underset{\color{#0066A7}{\hbox{\(\tfrac{du}{2}\)}}}{\underbrace{x\,dx}}=\int \sqrt{u}\,\dfrac{du}{2}=\dfrac{1}{2}\int u^{1/2}du=\dfrac{1}{2}\left(\dfrac{u^{3/2}}{\dfrac{3}{2}}\right) +C \\ &= & \dfrac{(x^{2}+1) ^{3/2}}{3} +C \end{eqnarray*} \]

(c) We let \(u=\sqrt{x}=x^{1/2}\). Then \(du=\dfrac{1}{2}x^{-1/2}dx=\dfrac{dx}{2\sqrt{x}}\), so \(\dfrac{dx}{\sqrt{x}}\) \(=2du\). \[ \int \dfrac{e^{\sqrt{{ x}}}}{\sqrt{x}}dx=\int e^{\sqrt{x}}\cdot \dfrac{dx}{\sqrt{x}}=\int e^{u}\cdot 2du=2e^{u}+C=2e^{\sqrt{{ x}}}+C \]