This web exercise gives practice with a variety of indefinite integration problems.
All the calculus steps are shown in the solutions.
In this exercise, all functions are assumed to have the required properties for a particular situation.
For example, any function in a denominator is assumed
to be nonzero, any function inside a logarithm is assumed to be positive, and so on.
In general, integration is more difficult than differentiation.
The following list of questions/considerations offers a thought process to help you
come up with an appropriate tool for handling a given integration problem:
IS IT JUST A DIFFERENTIATION RULE, IN REVERSE?
The first question you should always consider is:
Do I know a function whose
derivative is blah?
Is this just a differentiation formula,
‘in reverse’?
EXAMPLE:
$\displaystyle\int \sec\, x\,\tan\, x\ dx =\ ?$
Do you know a function whose derivative is $\,\sec\,x\,\tan\,x\,$?
Yes! Recall that $\frac{d}{dx} \sec\,x = \sec\, x\,\tan x\,$; so, $\,\sec x\,$ is an
antiderivative of $\,\sec\, x\, \tan\, x\,.$
Thus: $\displaystyle\int \sec\, x\,\tan\, x\ dx = \sec\,x + C$
EXAMPLE:
$\displaystyle\int \frac{1}{\sqrt{1-x^2}}\,dx =\ ?$
Do you know a function whose derivative is $\,\frac{1}{\sqrt{1-x^2}}\,$?
Yes! Recall that $\frac{d}{dx} \arcsin\,x = \frac{1}{\sqrt{1-x^2}}\,$; so, $\,\arcsin x\,$ is an
antiderivative of $\,\frac{1}{\sqrt{1-x^2}}\,.$
Thus: $\displaystyle\int \frac{1}{\sqrt{1-x^2}}\,dx =\ \arcsin\,x + C$
SOMETIMES, YOU JUST NEED TO RENAME..
Sometimes, the function being integrated needs to be re-named, using basic algebra skills.
Perhaps you need to write a
radical as a rational exponent.
Perhaps you need to use a distributive law.
Perhaps you need to FOIL something out.
Perhaps you need to use the fact that $\frac{A+B}C = \frac{A}{C} + \frac{B}{C}\,.$
EXAMPLE:
$\displaystyle\int\frac{1}{\root 3\of x}\ dx
= \int x^{-1/3}\ dx
= \frac{x^{-\frac 13 + 1}}{-\frac13 + 1} + C
= \frac{x^{2/3}}{2/3} + C
= \frac 32 \root 3\of{x^2} + C
$
If a problem starts off in radical form, then you should
give your answer in radical form.
Recall that dividing by a fraction is the same as multiplying by its reciprocal: e.g.,
dividing by $\,\frac 23\,$ is the same as multiplying by $\,\frac 32\,.$
Also recall that for $\,n\ne -1\,,$ $\,\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\,.$
EXAMPLE:
$\displaystyle\int \frac{x^3-1}{x}\ dx
= \int \bigl(\frac{x^3}x - \frac 1x\bigr)\ dx
= \int( x^2 - \frac 1x)\ dx
= \frac{x^3}3 - \ln |x| + C
$
SUBSTITUTION
There's a common theme in mathematics: if you can't solve a problem, try to ‘turn it into’
a problem that you can solve.
Solve the new problem, and then use this answer to solve the original problem.
Substitution is just such a method!
Key idea for substitution: look for something in the integrand whose derivative is also a factor
in the integrand, perhaps off by a constant.
This idea is illustrated in the examples below.
This technique is often called ‘$u$-substitution’ since $u$ is a common variable
used in the renaming process.
The expression chosen for $\,u\,$ is often something in parentheses, something in an exponent,
something in a denominator, or something under a radical.
EXAMPLE:
In the integral below, note that the derivative of $\,3x^2 - 1\,$ is $\,6x\,$; the variable part of $\,6x\,$ is $\,x\,.$
There is a factor of $\,x\,$ in the integrand! Woo hoo!
Note how we multiply by $\,1\,$ in the form of $\,\frac{6}{6} = 6\cdot\frac{1}{6}\ $:
the $\,6\,$ is left inside the integrand (where it becomes part of $\,du\,$)
but the $\,\frac{1}{6}\,$ is slid out of the integral.
Here, ‘$\ du = 6x\,dx\ $’ can be viewed as a convenient reformulation of $\displaystyle\ \frac{du}{dx} = 6x\ $.
Note also that we started with an integral in $\,x\,,$ so we must end up with an integral in $\,x\,.$
Let $\,\color{purple}{u = 3x^2-1}\,,$ so that $\color{red}{\,du = 6x\ dx\,}$. Then,
$\displaystyle\int x(3x^2 - 1)^7\ dx = \frac{1}{6}\int (\color{purple}{3x^2 - 1})^7\,(\color{red}{6x\,dx}) = \frac{1}{6}\int u^7\ du$
$\displaystyle = \frac{1}{6}\cdot \frac{u^8}{8} + C = \frac{(3x^2-1)^8}{48} + C$
EXAMPLE:
In the integral below, recall that $\,\sin^5(3x)\,$ is a shorthand for $\,(\sin(3x))^5\,.$
The derivative of $\,\sin(3x)\,$ is $\,3\cos(3x)\,$; the variable part is $\,\cos(3x)\,,$ which is a factor in the integrand.
Let
$\,u = \sin(3x)\,$, so that $\color{red}{\,du = 3\cos(3x)\ dx\,}$. Then,
$\displaystyle\int \sin^5(3x)\cos(3x)\ dx
= \frac{1}{3}\int (\color{purple}{\sin(3x)})^5 (\color{red}{3\cos 3x\ dx})
= \frac{1}{3}\int u^5\ du$
$\displaystyle = \frac{1}{3}\cdot \frac{u^6}{6} + C
= \frac{\sin^6(3x)}{18} + C$
INTEGRATION BY PARTS
Integration by parts is the integration counterpart to the product rule for differentiation.
Let $u$ and $v$ both be functions of $x$, so that:
$$\frac{d}{dx} uv = u\frac{dv}{dx} + v\frac{du}{dx}$$
Integrating both sides with respect to $\,x\,$ gives:
$$\int \frac{d}{dx} uv\ dx = \int u\frac{dv}{dx}\ dx + \int v\frac{du}{dx}\ dx$$
This simplifies to:
$$uv = \int u\,dv + \int v\,du$$
A bit of rearrangement gives the INTEGRATION BY PARTS FORMULA:
$$\int u\,dv = uv - \int v\,du$$
Here are some observations regarding this formula:
- Can't solve an integral by prior considerations? Then, try parts.
-
You must choose $\,u\,$ and $\,dv\ $:
- $u$ is differentiated to find $\,du\,$
- $\,dv\,$ is integrated to find $\,v\ $ (usually, the constant of integration is chosen to be zero)
-
Choose something for $\,dv\,$ that you know how to integrate (else you can't
find $\,v\,,$ and hence can't use the parts formula).
-
More than one choice for $\,dv\,$? Then, choose something
for $\,u\,$ that gets simpler when you integrate it.
EXAMPLE:
The classic problem that requires integration by parts is integrating the natural logarithm function.
Let $\,dv = dx\,,$ so that $\,v = x\,.$
Let $u = \ln x\,,$ so that $\,du = \frac 1x\,dx\,.$
Then,
$\int \overbrace{\ln x}^{u}\ \overbrace{\ dx\ }^{dv} =
\overbrace{(\ln x)}^{u}\cdot\overbrace{\ \ x\ \strut\ }^{v} - \int \overbrace{\strut\ \ x\ \ }^{v}\cdot\overbrace{\frac{1}{x}\,dx}^{du}
= x\ln x - \int (1)\,dx
= x\ln x - x + C$
EXAMPLE:
Let $u = x\,,$ so that $\,du = \,dx\,.$
Let $\,dv = {\text{e}}^x\,dx\,,$ so that $\,v = {\text{e}}^x\,.$
Then,
$\int x{\text{e}}^x\,dx
= x{\text{e}}^x - \int {\text{e}}^x\,dx
= x{\text{e}}^x - {\text{e}}^x + C
$
Master the ideas from this section
by practicing the exercise at the bottom of this page.