Integration by Substitution

Complicated integrals can sometimes be made simpler through the method of substitution. Substitution involves condensing an expression of $x$ into a single variable, say $u$, and then expressing the integral in terms of $u$ instead of $x$.

To make the idea of substitution more concrete, consider the integral $\int (3x+1)^8 \, dx$. We may be tempted to use the power rule, and say that the integral evaluates to $\frac{1}{9}(3x+1)^9$. But if we differentiate to check our result, we see that, because of the chain rule, the derivative of this expression is not equal to the function inside the integral.

$\begin{align*} \left[ \frac{1}{9}(3x+1)^9 \right]' &= \frac{1}{9} \left[ (3x+1)^9 \right]' \\ &= \frac{1}{9} \cdot 9(3x+1)^8 ( 3x+1 )' \\ &= (3x+1)^8 ( 3) \\ &= 3(3x+1)^8 \\ &\neq (3x+1)^8 \end{align*}$


To turn the integral into one that can be solved with the power rule, we condense the $3x+1$ expression into a single variable $u$, through the substitution $u=3x+1$.

$\begin{align*} \int (3x+1)^8 \, dx &= \int u^8 \, dx \end{align*}$


Before we apply the power rule, we need to take care of one issue: the differential is still $dx$, and we need it to be $du$. In general, we can’t just replace the $dx$ differential with a $du$ differential. However, by interpreting the derivative as a fraction, we can solve for the $dx$ differential in terms of the $du$ differential.

$\begin{align*} \frac{du}{dx} &= (3x+1)' \\ \frac{du}{dx} &= 3 \\ du &= 3dx \\ \frac{1}{3}du &= dx \\ dx &= \frac{1}{3} du \end{align*}$


Once our integral is fully expressed in terms of $u$, we can solve it via the power rule, and then substitute $u=3x+1$ again to write our answer in terms of $x$.

$\begin{align*} \int u^8 \, dx &= \int u^8 \cdot \frac{1}{3}\, du \\ &= \frac{1}{3} \int u^8 \, du \\ &= \frac{1}{3} \cdot \frac{1}{9}u^9 + C \\ &= \frac{1}{27}u^9 + C \\ &= \frac{1}{27}(3x+1)^9 + C \end{align*}$


We verify that the derivative of the result is indeed the original function within the integral.

$\begin{align*} \left[ \frac{1}{27}(3x+1)^9 \right]' &= \frac{1}{27} \left[ (3x+1)^9 \right]' \\ &= \frac{1}{27} \cdot 9(3x+1)^8 ( 3x+1 )' \\ &= \frac{1}{3} (3x+1)^8 ( 3) \\ &= (3x+1)^8 \end{align*}$


The key to substitution is choosing the right substitution. But how can we tell what is the right substitution? For example, in the integral below, should we substitute $u=\sin x$ or $u=\cos x$?

$\begin{align*} \int \sin^2x \cos x \, dx \end{align*}$


Whenever we are torn between multiple substitution choices, we should choose the substitution whose derivative will cancel out other terms in the integral. In this case, we should choose $u=\sin x$, because the derivative $u’=\cos x$ will cancel out the existing $\cos x$ inside the integral. On the other hand, would not work, because the derivative $u’=-\sin x$ would not fully cancel the existing $\sin^2 x$ inside the integral.

Choosing $u=\sin x$, we have $\frac{du}{dx} = \cos x$, so $dx = \frac{1}{\cos x} du$. Substituting into the integral, we are able to evaluate.

$\begin{align*} \int \sin^2 x \cos x \, dx &= \int u^2 \cos x \cdot \frac{1}{\cos x} du \\ &= \int u^2 du \\ &= \frac{1}{3}u^3 + C \\ &= \frac{1}{3} \sin^3 x + C \end{align*}$



Practice Problems


Evaluate each integral, using substitution. (You can view the solution by clicking on the problem.)

$\begin{align*} 1) \hspace{.5cm} \int \sqrt{x+2} \, dx \end{align*}$
Solution:
$\begin{align*} \frac{2}{3} (x+2)^\frac{3}{2} + C \end{align*}$


$\begin{align*} 2) \hspace{.5cm} \int (4x+3)^8 \, dx \end{align*}$
Solution:
$\begin{align*} \frac{1}{36} (4x+3)^9 + C \end{align*}$


$\begin{align*} 3) \hspace{.5cm} \int \frac{x}{\sqrt{1−x^2}} \, dx \end{align*}$
Solution:
$\begin{align*} −\sqrt{1−x^2} + C \end{align*}$


$\begin{align*} 4) \hspace{.5cm} \int \frac{3x^2}{(x^3−5)^3} \, dx \end{align*}$
Solution:
$\begin{align*} −\frac{1}{2(x^3−5)^2} + C \end{align*}$


$\begin{align*} 5) \hspace{.5cm} \int \sec^2 x \tan^2 x \, dx \end{align*}$
Solution:
$\begin{align*} \frac{1}{3} \tan^3 x + C \end{align*}$


$\begin{align*} 6) \hspace{.5cm} \int \frac{\cos x}{\sqrt{\sin x}} \, dx \end{align*}$
Solution:
$\begin{align*} 2\sqrt{\sin x} + C \end{align*}$


$\begin{align*} 7) \hspace{.5cm} \int \frac{\sec^2 x \tan x}{(\sec^2 x + 1)^4} \, dx \end{align*}$
Solution:
$\begin{align*} −\frac{1}{6(\sec^2 x + 1)^3} + C \end{align*}$


$\begin{align*} 8) \hspace{.5cm} \int \cos (\cos x) \sin x \, dx \end{align*}$
Solution:
$\begin{align*} −\sin (\cos x) + C \end{align*}$


$\begin{align*} 9) \hspace{.5cm} \int x^2 e^{x^3+1} \, dx \end{align*}$
Solution:
$\begin{align*} \frac{1}{3} e^{x^3+1} + C \end{align*}$


$\begin{align*} 10) \hspace{.5cm} \int \frac{\csc^2 x}{e^{\cot x}} \, dx \end{align*}$
Solution:
$\begin{align*} e^{−\cot x} + C \end{align*}$


$\begin{align*} 11) \hspace{.5cm} \int \frac{x^2 e^{\sqrt{x^3−1}} }{\sqrt{x^3−1} } \, dx \end{align*}$
Solution:
$\begin{align*} \frac{2}{3} e^{\sqrt{x^3−1} }+C \end{align*}$


$\begin{align*} 12) \hspace{.5cm} \int e^{x+e^x}\, dx \end{align*}$
Solution:
$\begin{align*} e^{e^x} + C \end{align*}$


$\begin{align*} 13) \hspace{.5cm} \int \frac{e^x}{1+e^{2x}} \, dx \end{align*}$
Solution:
$\begin{align*} \arctan (e^x) + C \end{align*}$


$\begin{align*} 14) \hspace{.5cm} \int \frac{1}{x \sqrt{1−(\ln x)^2} } \, dx \end{align*}$
Solution:
$\begin{align*} \arcsin (\ln x) + C \end{align*}$


$\begin{align*} 15) \hspace{.5cm} \int \frac{1}{\sqrt{x−x^2}}\, dx \end{align*}$
Solution:
$\begin{align*} 2\arcsin \sqrt{x} + C \end{align*}$


$\begin{align*} 16) \hspace{.5cm} \int \frac{x}{x^4+1} \, dx \end{align*}$
Solution:
$\begin{align*} \frac{1}{2} \arctan x^2 + C \end{align*}$


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