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:

$\displaystyle\int\frac{1}{\root 3\of x}\ dx = \int x^{-1/3}\ dx = \frac{x^{-\frac 13 + 1}}{-\frac13 + 1} + C = \frac{x^{2/3}}{2/3} + C = \frac 32 \root 3\of{x^2} + C$

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