Find:

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 $\dfrac{dx}{\sqrt{x}}$ $=2du$. $$\int \dfrac{e^{\sqrt{{ x}}}}{\sqrt{x}}dx=\int e^{\sqrt{x}}\cdot \dfrac{dx}{\sqrt{x}}=\int e^{u}\cdot 2du=2e^{u}+C=2e^{\sqrt{{ x}}}+C$$