# Definite Integration by Substitution

\large \begin{align} \displaystyle u &= f(x) \\ du &= \dfrac{du}{dx} \times dx \\ \end{align}

Converting $x$-values to corresponding $u$-values is required.

## Example 1

Find $\displaystyle 2\int_{0}^{1}{\sqrt{2x+1}}dx$.

\begin{align} \displaystyle \text{Let } u &= 2x+1 \\ \dfrac{du}{dx} &= 2 \\ du &= 2dx \\ u &= 2 \times 1 + 1 = 3 &x=1\\ u &= 2 \times 0 + 1 = 1 &x=0\\ 2 \int_{x=0}^{x=1}{\sqrt{2x+1}}dx &= \int_{x=0}^{x=1}{\sqrt{u}}2dx \\ &= \int_{u=1}^{u=3}{u^{\frac{1}{2}}}du \\ &= \dfrac{1}{\frac{1}{2}+1}\left[u^{\frac{1}{2}+1}\right]_{1}^{3} \\ &= \dfrac{2}{3}\left[u^{\frac{3}{2}}\right]_{1}^{3} \\ &= \dfrac{2}{3}\left[\sqrt{u^3}\right]_{1}^{3} \\ &= \dfrac{2}{3}\left[\sqrt{3^3}-\sqrt{1^3}\right] \\ &= \dfrac{2}{3}\left[\sqrt{27}-1\right] \end{align}

It is not required to convert $x$-values to corresponding $u$-values, but to substitute $x$-values after integrating it.

## Example 2

Find $\displaystyle \int_{0}^{1}{x\sqrt{1+x^2}}dx$.

\begin{align} \displaystyle \text{Let } u &= 1+x^2 \\ \dfrac{du}{dx} &= 2x \\ du &= 2xdx \\ \int_{x=0}^{x=1}{x\sqrt{1+x^2}}dx &= \dfrac{1}{2}\int_{x=0}^{x=1}{\sqrt{1+x^2}}2xdx \\ &= \dfrac{1}{2}\int_{x=0}^{x=1}{\sqrt{u}}du \\ &= \dfrac{1}{2}\int_{x=0}^{x=1}{u^{\frac{1}{2}}}du \\ &= \dfrac{1}{2} \times \dfrac{1}{\frac{1}{2}+1}\left[u^{\frac{1}{2}+1}\right]_{x=0}^{x=1} \\ &= \dfrac{1}{2} \times \dfrac{1}{\frac{3}{2}}\left[u^{\frac{3}{2}}\right]_{x=0}^{x=1} \\ &= \dfrac{1}{2} \times \dfrac{2}{3}\left[u^{\frac{3}{2}}\right]_{x=0}^{x=1} \\ &= \dfrac{1}{3} \left[\sqrt{u^3}\right]_{x=0}^{x=1} \\ &= \dfrac{1}{3} \left[\sqrt{(1+x^2)^3}\right]_{x=0}^{x=1} \\ &= \dfrac{1}{3} \sqrt{(1+1^2)^3}-\sqrt{(1+0^2)^3} \\ &= \dfrac{\sqrt{8}-\sqrt{1}}{3} \\ &= \dfrac{\sqrt{8}-1}{3} \end{align}

## Example 3

Find $\displaystyle \int_{0}^{1}{(x^2+3x)^4(2x+3)}dx$.

\begin{align} \displaystyle \text{Let } u &= x^2+3x \\ \dfrac{du}{dx} &= 2x+3 \\ du &= (2x+3)dx \\ u &= 1^2 + 3 \times 1 = 4 &x=1 \\ u &= 0^2 + 3 \times 1 = 3 &x=0 \\ \int_{x=0}^{x=1}{(x^2+3x)^4(2x+3)}dx &= \int_{u=4}^{u=3}{u^4}du \\ &= \dfrac{1}{4+1}\left[u^{4+1}\right]_{3}^{4} \\ &= \dfrac{1}{5}\left[u^{5}\right]_{3}^{4} \\ &= \dfrac{1}{5}\left[4^{5}-3^{5}\right] \\ &= \dfrac{781}{5} \\ \end{align}