Integration by substitution

For $\int 2x(x^2+3)^3\,dx$, set the inside of the brackets to $t=x^2+3$. Then $\frac{dt}{dx}=2x$, so $dt=2x\,dx$, and $2x\,dx$ is replaced by $dt$.

The steps are: choose $t$, find $dt$ and remove $dx$, integrate in $t$, and substitute back to $x$. For expressions with $\sqrt{1-x^2}$, the substitution $x=\sin\theta$ is also used.

Examples

1. 01Which one can be solved with the substitution $t=x^{2}+1$?$\displaystyle$A$\displaystyle \int 2x(x^{2}+1)^{3}\,dx$B$\displaystyle \int \frac{1}{x^{2}+1}\,dx$C$\displaystyle \int 2x\,dx$D$\displaystyle \int x^{2}(x+1)^{3}\,dx$AnswerHide

   Answer$\displaystyle \int 2x(x^{2}+1)^{3}\,dx$

   Steps

   1. Substitution works when the derivative of the inside is already multiplied in. Differentiating $t$ gives
   2. $\displaystyle t$$\displaystyle = x^{2}+1$$\displaystyle \frac{dt}{dx}$$\displaystyle = 2x$
   3. $\int 2x(x^{2}+1)^{3}\,dx$ has that $2x$, so it turns into $dt$. The others do not, so nothing turns into $dt$
2. 02In $dt = \square\,dx$, what goes in □?$\displaystyle \begin{gathered} \displaystyle \int 2x(x^{2}+3)^{3}\,dx, \\[6pt] \displaystyle t=x^{2}+3 \end{gathered}$AnswerHide

   Answer$\displaystyle 2x$

   Steps

   1. Differentiate the $t$ we set, with respect to $x$
   2. $\displaystyle t$$\displaystyle = x^{2}+3$$\displaystyle \frac{dt}{dx}$$\displaystyle = 2x$
   3. That $2x$ is what $dt = 2x\,dx$ means; the constant $3$ vanishes when differentiated
3. 03Substitute back to $x$:$\displaystyle \begin{gathered} \displaystyle \int 2x(x^{2}+3)^{3}\,dx = \frac{t^{4}}{4}+C, \\[6pt] \displaystyle t=x^{2}+3 \end{gathered}$AnswerHide

   Answer$\displaystyle \frac{(x^{2}+3)^{4}}{4}+C$

   Steps

   1. Just put back what $t$ stood for
   2. $\displaystyle t$$\displaystyle = x^{2}+3$$\displaystyle \frac{t^{4}}{4}$$\displaystyle = \frac{(x^{2}+3)^{4}}{4}$
   3. The power was on $t$, so it lands on the whole bracket. Do not multiply by $2x$; it was already used up as $dt$
4. 04Rewrite as an integral in $t$:$\displaystyle \begin{gathered} \displaystyle \int 2x(x^{2}+2)^{3}\,dx, \\[6pt] \displaystyle t=x^{2}+2,\; dt=2x\,dx \end{gathered}$AnswerHide

   Answer$\displaystyle \int t^{3}\,dt$

   Steps

   1. $(x^{2}+2)^{3}$ becomes $t^{3}$, and the whole $2x\,dx$ becomes $dt$
   2. The $2x$ is absorbed into $dt$, so it does not stay in front; and neither does $dx$
   3. $\displaystyle \int t^{3}\,dt$$\displaystyle = \frac{t^{4}}{4}+C$
5. 05In $dt = \square\,dx$, what goes in □?$\displaystyle \begin{gathered} \displaystyle \int 2x(x^{2}+4)^{3}\,dx, \\[6pt] \displaystyle t=x^{2}+4 \end{gathered}$AnswerHide

   Answer$\displaystyle 2x$

   Steps

   1. Differentiate the $t$ we set, with respect to $x$
   2. $\displaystyle t$$\displaystyle = x^{2}+4$$\displaystyle \frac{dt}{dx}$$\displaystyle = 2x$
   3. That $2x$ is what $dt = 2x\,dx$ means; the constant $4$ vanishes when differentiated
6. 06Rewrite as an integral in $t$:$\displaystyle \begin{gathered} \displaystyle \int 2x(x^{2}+3)^{4}\,dx, \\[6pt] \displaystyle t=x^{2}+3,\; dt=2x\,dx \end{gathered}$AnswerHide

   Answer$\displaystyle \int t^{4}\,dt$

   Steps

   1. $(x^{2}+3)^{4}$ becomes $t^{4}$, and the whole $2x\,dx$ becomes $dt$
   2. The $2x$ is absorbed into $dt$, so it does not stay in front; and neither does $dx$
   3. $\displaystyle \int t^{4}\,dt$$\displaystyle = \frac{t^{5}}{5}+C$

FAQ

Q1What should I set as $t$?AnswerHide

A

Choose the part whose derivative appears as a factor outside. In $2x(x^2+3)^3$, the factor $2x$ is the derivative of $x^2+3$, so set $t=x^2+3$.

Q2Do I have to substitute back to $x$?AnswerHide

A

For an indefinite integral, yes: the answer is written in $x$, so replace $t$ with $x^2+3$. For a definite integral, you can change the limits to values of $t$ instead and skip substituting back.