When using the substitution method with definite integrals, the limits of integration must be rewritten in terms of the new variable.

Examples

Find   \( \int_{0}^2\ x(2+3x)^4\ dx \)   using the substitution   \(u=2+3x \)

\begin{align} Let \quad I & = \int_{0}^{2}\ x(2+3x)^4 \ dx \\ \\ And\ let \quad u & = 2+3x \\ \\ & This\ implies\ that\ our\ new\ limits\ are \\ \\ u & =2+3(0) =2 \\ and \quad u & = 2+3(2) =8 \\ \\ Also\ that \quad x & = \frac{u-2}{3} \\ \\ and \quad \frac{du}{dx} & = 3 \\ dx & = \frac{1}{3}du \\ \\ By\ substitution\ \quad I & = \int_{2}^{8} \ \left(\frac{u-2}{3} \right)u^4\ \frac{1}{3}\ du \\ \\ & = \int_{2}^{8} \frac{u^5-2u^4}{9}\ du \\ \\ & = \frac{1}{9} \int_{2}^{8} \left(u^5-2u^4 \right) \ du \\ \\ & = \frac{1}{9} \left[ \frac{u^6}{6}- \frac{2u^5}{5} \right]_{2}^{8} \\ \\ & = \frac{1}{9} \left[ \left( \frac{8^6}{6} - \frac{2 \left( 8^5 \right) }{5} \right) - \left( \frac{2^6}{6} - \frac{2 \left( 2^5 \right) }{5} \right) \right] \\ \\ & = \frac{16992}{5} \\ \\ \end{align}

Find   \( \int_{0}^{\frac{\pi}{2}} sinx \sqrt{2cosx+3}\ dx \)   using the substitution   \(u=cosx\)

\begin{align} Let \quad I & = \int_{0}^{\frac{\pi}{2}} sinx \sqrt{2cosx+3}\ dx \\ \\ And \quad let \quad u & = cosx \\ \\ This\ implies\ that\ our\ & new\ limits\ are \\ \\ u & =cos(0) =1 \\ and \quad u & = cos \left(\frac{\pi}{2} \right) =0 \\ \\ Also\ that \quad \frac{du}{dx} & =-sinx \\ dx & = -\frac{1}{sinx}\ du \\ \\ So \quad I & = \int_{1}^{0} \ sinx\ \sqrt{2u+3} \ \left(-\frac{1}{sinx} \right) du \\ \\ & = \int_{0}^{1} \ (2u+3)^{\frac{1}{2}}\ du \\ \\ & = \left[\frac{1}{3}(2u+3)^{\frac{3}{2}} \right]_{0}^1 \\ \\ & = \frac{ 5^{\frac{3}{2}} - 3^{\frac {3}{2}}}{3} \\ \\ \end{align}