Integration by Substitution

Find   \( \int\ x(3+2x)^4\ dx \)   using the substitution   \(u=3+2x \)

\begin{align} Let \quad I & = \int\ x(3+2x)^4 \ dx \\ \\ And\ let \quad u & = 3+2x \\ \\ Then \quad x & = \frac{u-3}{2} \\ \\ and \quad \frac{du}{dx} & = 2 \\ dx & = \frac{1}{2}du \\ \\ So \quad I & = \int \ \left(\frac{u-3}{2} \right)u^4\ \frac{1}{2}\ du \\ \\ & = \int \frac{u^5-3u^4}{4}\ du \\ \\ & = \int \left(\frac{u^5}{4}- \frac{3u^4}{4} \right) \ du \\ \\ & = \frac{1}{24}u^6 - \frac{3}{20}u^5 +c \\ \\ & = \frac{(3+2x)^6}{24} - \frac{3(3+2x)^5}{20} +c \end{align}

Find   \( \int sinx \sqrt{4cosx+2}\ dx \)   using the substitution   \(u=cosx\)

\begin{align} Let \quad I & = \int sinx \sqrt{4cosx+2}\ dx \\ \\ And \quad let \quad u & = cosx \\ \\ then \quad \frac{du}{dx} & =-sinx \\ dx & = -\frac{1}{sinx}\ du \\ \\ So \quad I & = \int \ sinx\ \sqrt{4u+2} \ \left(-\frac{1}{sinx} \right) du \\ \\ & = -\int \ (4u+2)^{\frac{1}{2}}\ du \\ \\ & = - \frac{1}{6}(4u+2)^{\frac{3}{2}}+c \\ \\ & = - \frac{1}{6}(4cosx+2)^{\frac{3}{2}} +c \end{align}