﻿ Mixed Integration Practice
MIXED INTEGRATION PRACTICE
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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.

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
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