Module 3: Integration By Substitution

The chain rule says that

$\frac{d F (G (x))}{d x} = F^{'} (G (x)) \cdot G^{'} (x),$ $\frac{dF(G(x))}{dx}=F'(G(x))\cdot G'(x),$

so that

$\int F^{'} (G (x)) \cdot G^{'} (x) d x = F (G (x)) + C .$ $\int F'(G(x))\cdot G'(x) dx= F(G(x))+C.$

We are going to rewrite this with notation you may be more familiar with.
Let's set LaTeX: u=G(x) $u=G(x)$ , and agree to write

$d u = G^{'} (x) d x$ $du=G'(x)dx$

$\int F^{'} (G (x)) G^{'} (x) d x = \int F^{'} (u) d u = F (u) + C = F (G (x)) + C$ $\int F'(G(x))G'(x) \; dx=\int F'(u)du=F(u)+C=F(G(x))+C$

2.1 A Few Examples

1. Compute $\int (1 - 3 x)^{47} d x$ $\int (1-3x)^{47}dx$ This integral currently does not have the form of any of the basic integrals that we mentioned in the introduction section.

If we replace $1 - 3 x$ $1-3x$ by $u$ $u$ , we have:

$u = 1 - 3 x \Rightarrow d u = (- 3) d x \Rightarrow d x = \frac{d u}{- 3}$ $u=1-3x \Rightarrow du=(-3)dx \Rightarrow dx=\frac{du}{-3}$

$\int (1 - 3 x)^{47} d x = \int u^{47} \frac{d u}{- 3} = \frac{- 1}{3} \int u^{47} d u$ $\int (1-3x)^{47}dx=\int u^{47}\frac{du}{-3}=\frac{-1}{3}\int u^{47}du$

$\int u^{47} d u$ $\int u^{47}du$ is in the form we know how to compute.

$\int (1 - 3 x)^{47} d x = \frac{- 1}{3} \int u^{47} d u = \frac{- 1}{3} \cdot \frac{1}{48} u^{48} + C = \frac{- 1}{3} \cdot \frac{1}{48} (1 - 3 x)^{48} + C$ $\int (1-3x)^{47}dx=\frac{-1}{3}\int u^{47}du=\frac{-1}{3}\cdot\frac{1}{48}u^{48}+C=\frac{-1}{3}\cdot\frac{1}{48}(1-3x)^{48}+C\$

2. Compute $\int \frac{\ln x}{x} d x$ $\int \frac{\ln x}{x} \; dx$ .

If we set $u = \ln x$ $u=\ln x$ , then we will have $d u = \frac{1}{x} d x$ $du=\frac{1}{x}dx$ and $d x = x d u$ $dx=xdu$ . Therefore,

$\int \frac{\ln x}{x} d x = \int \frac{u}{x} x d u = \int u d u = \frac{1}{2} u^{2} + C = \frac{1}{2} \ln^{2} (x) + C$ $\int \frac{\ln x}{x} \; dx=\int \frac{u}{x}\;x\;du=\int u\; du=\frac{1}{2}u^2+C=\frac{1}{2}\ln^2(x)+C\$

3. Compute $\int ((x^{2} - 1) (x + 1)^{\frac{- 2}{3}}) d x$ $\int \left((x^2-1)(x+1)^{\frac{-2}{3}} \right)\; dx$ .

a) Here is one approach. Let's first do some algebraic manipulation and rewrite the integral.

$\begin{aligned} \int ((x^{2} - 1) (x + 1)^{\frac{- 2}{3}}) d x & = \int ((x - 1) (x + 1) (x + 1)^{\frac{- 2}{3}}) d x \\ = \int ((x - 1) (x + 1)^{\frac{1}{3}}) d x \end{aligned}$ $\begin{align*} \int \left((x^2-1)(x+1)^{\frac{-2}{3}} \right)\; dx &= \int \left((x-1)(x+1)(x+1)^{\frac{-2}{3}} \right)\; dx\\ &= \int \left((x-1)(x+1)^{\frac{1}{3}} \right)\; dx\\ \end{align*}$

If we let $u = x - 1$ $u=x-1$ then $d u = d x$ $du=dx$ and $x + 1 = u + 2$ $x+1 = u + 2$ .

$\begin{aligned} \int ((x - 1) (x + 1)^{\frac{1}{3}}) d x & = \int u (u + 2)^{\frac{1}{3}} d u . \end{aligned}$ $\begin{align*} \int \left((x-1)(x+1)^{\frac{1}{3}} \right)\; dx&=\int u(u+2)^{\frac{1}{3}} \; du.\\ \end{align*}$

Let $v = u + 2$ $v=u+2$ then $d v = d u$ $dv=du$ and $u = v - 2.$ $u=v-2.$

$\begin{aligned} \int ((x - 1) (x + 1)^{\frac{1}{3}}) d x & = \int u (u + 2)^{\frac{1}{3}} d u . \\ = \int (v - 2) v^{\frac{1}{3}} d v \\ = \int v^{\frac{4}{3}} - 2 v^{\frac{1}{3}} d v \\ = \frac{1}{\frac{4}{3} + 1} v^{\frac{4}{3} + 1} - \frac{2}{\frac{1}{3} + 1} v^{\frac{1}{3} + 1} + C \\ = \frac{1}{\frac{4}{3} + 1} (u + 2)^{\frac{4}{3} + 1} - \frac{2}{\frac{1}{3} + 1} (u + 2)^{\frac{1}{3} + 1} + C \\ = \frac{1}{\frac{4}{3} + 1} (x + 1)^{\frac{4}{3} + 1} - \frac{2}{\frac{1}{3} + 1} (x + 1)^{\frac{1}{3} + 1} + C \end{aligned}$ $\begin{align*} \int \left((x-1)(x+1)^{\frac{1}{3}} \right)\; dx&=\int u(u+2)^{\frac{1}{3}} \; du.\\ &=\int (v-2)v^{\frac{1}{3}} \; dv\\ &=\int v^{\frac{4}{3}}-2v^{\frac{1}{3}}\; dv\\ &=\frac{1}{\frac{4}{3}+1}v^{\frac{4}{3}+1}-\frac{2}{\frac{1}{3}+1}v^{\frac{1}{3}+1}+C\\ &=\frac{1}{\frac{4}{3}+1}(u+2)^{\frac{4}{3}+1}-\frac{2}{\frac{1}{3}+1}(u+2)^{\frac{1}{3}+1}+C\\ &=\frac{1}{\frac{4}{3}+1}(x+1)^{\frac{4}{3}+1}-\frac{2}{\frac{1}{3}+1}(x+1)^{\frac{1}{3}+1}+C\\ \end{align*}$

Remark: The power of $\frac{1}{3}$ $\frac{1}{3}$ of $(x + 1)$ $(x+1)$ added complexity to our solution and required us to use substitution twice. Check out the next approach.

b) If we let $u = x + 1$ $u=x+1$ then $d u = d x$ $du=dx$ and $x - 1 = u - 2.$ $x-1=u-2.$

$\begin{aligned} \int ((x - 1) (x + 1)^{\frac{1}{3}}) d x & = \int (u - 2) (u)^{\frac{1}{3}} d u \\ = \int u^{\frac{4}{3}} - 2 u^{\frac{1}{3}} d u \\ = \frac{1}{\frac{4}{3} + 1} u^{\frac{4}{3} + 1} - \frac{2}{\frac{1}{3} + 1} u^{\frac{1}{3} + 1} + C \\ = \frac{1}{\frac{4}{3} + 1} (x + 1)^{\frac{4}{3} + 1} - \frac{2}{\frac{1}{3} + 1} (x + 1)^{\frac{1}{3} + 1} + C \end{aligned}$ $\begin{align*} \int \left((x-1)(x+1)^{\frac{1}{3}} \right)\; dx&=\int (u-2)(u)^{\frac{1}{3}} \; du\\ &=\int u^{\frac{4}{3}}-2u^{\frac{1}{3}}\; du\\ &=\frac{1}{\frac{4}{3}+1}u^{\frac{4}{3}+1}-\frac{2}{\frac{1}{3}+1}u^{\frac{1}{3}+1}+C\\ &=\frac{1}{\frac{4}{3}+1}(x+1)^{\frac{4}{3}+1}-\frac{2}{\frac{1}{3}+1}(x+1)^{\frac{1}{3}+1}+C\\ \end{align*}$

Remark: Both approaches are equally valid and correct. One has fewer steps. How do we decide on the choice of $u$ $u$ ? It takes practice, experience, and not being afraid of trying different options. Check out the next approach for the same problem!

c) If $u = \frac{1}{x + 1}$ $u=\frac{1}{x+1}$ then $d u = (- 1) (1 + x)^{- 2} d x = - u^{2} d x$ $du=(-1)(1+x)^{-2}dx=-u^2\;dx$ . Hence $d x = - \frac{1}{u^{2}} d u$ $dx=-\frac{1}{u^2}\; du$ .

$\begin{aligned} \int ((x^{2} - 1) (x + 1)^{\frac{- 2}{3}}) d x & = \int (x - 1) (x + 1)^{\frac{1}{3}} d x \\ = \int (\frac{1}{u} - 2) \frac{1}{u^{\frac{1}{3}}} (- \frac{1}{u^{2}}) d u \\ = - \int (u^{\frac{- 10}{3}} - 2 u^{\frac{- 7}{3}}) d u \\ = - ((\frac{1}{\frac{- 10}{3} + 1}) u^{\frac{- 10}{3} + 1} - 2 (\frac{1}{\frac{- 7}{3} + 1}) u^{\frac{- 7}{3} + 1}) + C \\ = - ((\frac{1}{\frac{- 10}{3} + 1}) {(\frac{1}{x + 1})}^{\frac{- 10}{3} + 1} - 2 (\frac{1}{\frac{- 7}{3} + 1}) {(\frac{1}{x + 1})}^{\frac{- 7}{3} + 1}) + C \end{aligned}$ $\begin{align*} \int \left((x^2-1)(x+1)^{\frac{-2}{3}} \right)\; dx&=\int (x-1)(x+1)^{\frac{1}{3}}\;dx\\ &=\int\left(\frac{1}{u}-2\right)\frac{1}{u^{\frac{1}{3}}} \left(-\frac{1}{u^2}\right)\; du\\ &=-\int\left( u^{\frac{-10}{3}}-2u^{\frac{-7}{3}}\right)du\\ &=-\left(\left(\frac{1}{\frac{-10}{3}+1}\right)u^{\frac{-10}{3}+1}-2 \left(\frac{1}{\frac{-7}{3}+1}\right)u^{\frac{-7}{3}+1}\right)+C\\ &=-\left(\left(\frac{1}{\frac{-10}{3}+1}\right)\left(\frac{1}{x+1}\right)^{\frac{-10}{3}+1}-2\left(\frac{1}{\frac{-7}{3}+1}\right)\left(\frac{1}{x+1}\right)^{\frac{-7}{3}+1}\right)+C\\ \end{align*}$

Remark: We do NOT need any further simplification. However, in order to convince ourselves that the third approach gives us the same answer as the first two tries, let's do some algebraic manipulations.

$\begin{aligned} - ((\frac{1}{\frac{- 10}{3} + 1}) {(\frac{1}{x + 1})}^{\frac{- 10}{3} + 1} - 2 (\frac{1}{\frac{- 7}{3} + 1}) {(\frac{1}{x + 1})}^{\frac{- 7}{3} + 1}) + C & = \\ - ((\frac{1}{\frac{- 7}{3}}) {(\frac{1}{x + 1})}^{\frac{- 7}{3}} - 2 (\frac{1}{\frac{- 4}{3}}) {(\frac{1}{x + 1})}^{\frac{- 4}{3}}) + C & = \\ \frac{1}{\frac{7}{3}} (x + 1)^{\frac{7}{3}} - 2 \frac{1}{\frac{4}{3}} (x + 1)^{\frac{4}{3}} + C & = \\ \frac{1}{\frac{4}{3} + 1} (x + 1)^{\frac{4}{3} + 1} - \frac{2}{\frac{1}{3} + 1} (x + 1)^{\frac{1}{3} + 1} + C . \end{aligned}$ $\begin{align*} -\left(\left(\frac{1}{\frac{-10}{3}+1}\right)\left(\frac{1}{x+1}\right)^{\frac{-10}{3}+1}-2\left(\frac{1}{\frac{-7}{3}+1}\right)\left(\frac{1}{x+1}\right)^{\frac{-7}{3}+1}\right)+C&=\\ -\left(\left(\frac{1}{\frac{-7}{3}}\right)\left(\frac{1}{x+1}\right)^{\frac{-7}{3}}-2\left(\frac{1}{\frac{-4}{3}}\right)\left(\frac{1}{x+1}\right)^{\frac{-4}{3}}\right)+C&=\\ \frac{1}{\frac{7}{3}}(x+1)^{\frac{7}{3}}-2\frac{1}{\frac{4}{3}}(x+1)^{\frac{4}{3}}+C&=\\ \frac{1}{\frac{4}{3}+1}(x+1)^{\frac{4}{3}+1}-\frac{2}{\frac{1}{3}+1}(x+1)^{\frac{1}{3}+1}+C.&\\ \end{align*}$

Remark: Let's reemphasize that this algebraic manipulations was not needed to compute the integral. This third approach was chosen to show, there are multiple ways of doing a $u -$ $u-$ substitution problem and some of them are not obvious at all, like the last one.

4. Compute $\int \frac{\sin (2 x)}{\sqrt{1 + \cos (2 x)}} d x$ $\int \frac{\sin(2x)}{\sqrt{1+\cos(2x)}}\; dx$

a) We let $u = s i n (2 x)$ $u=sin(2x)$ , then $d u = 2 c o s (2 x) d x$ $du=2cos(2x)dx$

$\begin{aligned} \int \frac{\sin (2 x)}{\sqrt{1 + \cos (2 x)}} d x & = \int \frac{u}{\sqrt{1 + \cos (2 x)}} \frac{d u}{2 \cos (2 x)} \end{aligned}$ $\begin{align*} \int \frac{\sin(2x)}{\sqrt{1+\cos(2x)}}\; dx&= \int \frac{u}{\sqrt{1+\cos(2x)}}\frac{du}{2\cos(2x)} \end{align*}$

It is hard to imagine what to do next to get rid of $x$ $x$ . So we will try another approach!

b) We let $u = c o s (2 x),$ $u=cos(2x),$ then $d u = - 2 s i n (2 x) d x .$ $du=-2sin(2x)dx.$

$\begin{aligned} \int \frac{\sin (2 x)}{\sqrt{1 + \cos (2 x)}} d x & = \frac{- 1}{2} \int \frac{- 2 \sin (2 x) d x}{\sqrt{1 + \cos (2 x)}} \\ = \frac{- 1}{2} \int \frac{d u}{\sqrt{1 + u}} \end{aligned}$ $\begin{align*} \int \frac{\sin(2x)}{\sqrt{1+\cos(2x)}}\; dx&= \frac{-1}{2}\int \frac{-2\sin(2x)\; dx}{\sqrt{1+\cos(2x)}}\\ &=\frac{-1}{2} \int \frac{du}{\sqrt{1+u}} \end{align*}$

The expression on the line above is not exactly of the forms we know. So once again we use substitution. We let $v = 1 + u$ $v=1+u$ . Then $d v = d u$ $dv=du$ and we have

$\begin{aligned} \int \frac{\sin (2 x)}{\sqrt{1 + \cos (2 x)}} d x & = \frac{- 1}{2} \int \frac{1}{\sqrt{1 + u}} d u \\ = \frac{- 1}{2} \int \frac{1}{\sqrt{v}} d v \\ = \frac{- 1}{2} \int v^{\frac{- 1}{2}} d v \\ = (\frac{- 1}{2}) (\frac{1}{\frac{- 1}{2} + 1}) v^{\frac{- 1}{2} + 1} + C \\ = (\frac{- 1}{2}) (\frac{1}{\frac{- 1}{2} + 1}) (1 + u)^{\frac{- 1}{2} + 1} + C \\ = (\frac{- 1}{2}) (\frac{1}{\frac{- 1}{2} + 1}) (1 + \cos (2 x))^{\frac{- 1}{2} + 1} + C \end{aligned}$ $\begin{align*} \int \frac{\sin(2x)}{\sqrt{1+\cos(2x)}}\; dx&= \frac{-1}{2}\int \frac{1}{\sqrt{1+u}}\; du\\ &=\frac{-1}{2}\int \frac{1}{\sqrt{v}}\; dv\\ &=\frac{-1}{2}\int v^{\frac{-1}{2}}\; dv\\ &=\left(\frac{-1}{2}\right)\left(\frac{1}{\frac{-1}{2}+1}\right)v^{\frac{-1}{2}+1}+C\\ &=\left(\frac{-1}{2}\right)\left(\frac{1}{\frac{-1}{2}+1}\right)(1+u)^{\frac{-1}{2}+1}+C\\ &=\left(\frac{-1}{2}\right)\left(\frac{1}{\frac{-1}{2}+1}\right)(1+\cos(2x))^{\frac{-1}{2}+1}+C\\ \end{align*}$

With a bit of reflection on this solution, we may notice a streamlined solution as described below.

c) We let $u = 1 + \cos (2 x)$ $u=1+\cos(2x)$ . Then $d u = - 2 \sin (2 x) d x$ $du=-2\sin(2x) dx$ .

$\begin{aligned} \int \frac{\sin (2 x)}{\sqrt{1 + \cos (2 x)}} d x & = \int \frac{\sin (2 x)}{\sqrt{u}} \frac{- d u}{2 \sin (2 x)} \\ = \frac{- 1}{2} \int \frac{1}{\sqrt{u}} d u \\ = \frac{- 1}{2} \int u^{\frac{- 1}{2}} d u \\ = (\frac{- 1}{2}) (\frac{1}{\frac{- 1}{2} + 1}) u^{\frac{- 1}{2} + 1} + C \\ = (\frac{- 1}{2}) (\frac{1}{\frac{- 1}{2} + 1}) (1 + \cos (2 x))^{\frac{- 1}{2} + 1} + C \end{aligned}$ $\begin{align*} \int \frac{\sin(2x)}{\sqrt{1+\cos(2x)}}\; dx&= \int \frac{\sin(2x)}{\sqrt{u}}\frac{-du}{2\sin(2x)}\\ &=\frac{-1}{2}\int \frac{1}{\sqrt{u}}\; du\\ %&=\frac{-1}{2}\int \frac{1}{\sqrt{v}}\; dv\\ &=\frac{-1}{2}\int u^{\frac{-1}{2}}\; du\\ &=\left(\frac{-1}{2}\right)\left(\frac{1}{\frac{-1}{2}+1}\right)u^{\frac{-1}{2}+1}+C\\ %&=\left(\frac{-1}{2}\right)\left(\frac{1}{\frac{-1}{2}+1}\right)(1+u)^{\frac{-1}{2}+1}+C\\ &=\left(\frac{-1}{2}\right)\left(\frac{1}{\frac{-1}{2}+1}\right)(1+\cos(2x))^{\frac{-1}{2}+1}+C\\ \end{align*}$