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### Course: AP®︎/College Calculus AB>Unit 6

Lesson 9: Finding antiderivatives and indefinite integrals: basic rules and notation: common indefinite integrals

# Common integrals review

Review the integration rules for all the common function types.

## Polynomials

$\int {x}^{n}\phantom{\rule{0.167em}{0ex}}dx=\frac{{x}^{n+1}}{n+1}+C$

$\begin{array}{rl}\int \sqrt[m]{\phantom{A}{x}^{n}}\phantom{\rule{0.167em}{0ex}}dx& =\int {x}^{{}^{\frac{n}{m}}}\phantom{\rule{0.167em}{0ex}}dx\\ \\ & =\frac{{x}^{{}^{\frac{n}{m}+1}}}{\frac{n}{m}+1}+C\end{array}$
Want to practice integrating polynomials and radicals? Check out these exercises:

## Trigonometric functions

$\int \mathrm{sin}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=-\mathrm{cos}\left(x\right)+C$
$\int \mathrm{cos}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=\mathrm{sin}\left(x\right)+C$
$\int {\mathrm{sec}}^{2}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=\mathrm{tan}\left(x\right)+C$
$\int {\mathrm{csc}}^{2}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=-\mathrm{cot}\left(x\right)+C$
$\int \mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=\mathrm{sec}\left(x\right)+C$
$\int \mathrm{csc}\left(x\right)\mathrm{cot}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=-\mathrm{csc}\left(x\right)+C$
Want to practice integrating trigonometric functions? Check out these exercises:

## Exponential functions

$\int {e}^{x}\phantom{\rule{0.167em}{0ex}}dx={e}^{x}+C$
$\int {a}^{x}\phantom{\rule{0.167em}{0ex}}dx=\frac{{a}^{x}}{\mathrm{ln}\left(a\right)}+C$

## Integrals that are logarithmic functions

$\int \frac{1}{x}\phantom{\rule{0.167em}{0ex}}dx=\mathrm{ln}|x|+C$
Want to learn more about integrating exponential functions and $\frac{1}{x}$? Check out this video.
Want to practice integrating exponential functions and $\frac{1}{x}$? Check out this exercise.

## Integrals that are inverse trigonometric functions

$\int \frac{1}{\sqrt{{a}^{2}-{x}^{2}}}\phantom{\rule{0.167em}{0ex}}dx=\mathrm{arcsin}\left(\frac{x}{a}\right)+C$
$\int \frac{1}{{a}^{2}+{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx=\frac{1}{a}\mathrm{arctan}\left(\frac{x}{a}\right)+C$

## Want to join the conversation?

• Why isn't there an arccos integral function?
• Basically, because the algebra doesn't work out nicely. There are plenty of derivatives of trig functions that exist, but there are only a few that result in a non-trig-function-involving equation.
For example, the derivative of arcsin(x/a)+c = 1/sqrt(a^2-x^2), doesn't involve any trig functions in it's derivative. If we reverse this process on 1/sqrt(a^2-x^2) (find the indefinite integral) we get arcsin(x/a)+C, so we went from an equation with no trig functions to an equation with trig functions.

There aren't many other equations that work out this nicely.
https://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integrands_involving_only_cosine
• Where is the video for the second function under "Exponential Functions" (the integral of a^x)? Where are the videos for the whole section of "Integrals that are inverse trigonometric functions" (the integral of 1/sqrt[(a^2)-(x^2)] and the integral of 1/sqrt[(a^2)+(x^2)]? I can't find these three functions mentioned anywhere in the videos.
• All these integrals of trigonometric functions are really confusing for me. Do I have to just learn them by heart? Or is there some section I missed, where they are explained more intuitively?
• There are proofs out there for each trig function but it is much easier to just learn them by heart.
• Why is the integral of tan(x) not listed?
• Any mnemonics to remember these? Anyone?
• Just commit the derivatives to memory and then use the opposites to remember these!
• Could someone please provide me with the proof for
integral of 1/(a^2 + x^2)
• 1/(a² + x²) = 1/(a²(1 + x²/a²)
Let x = a·tan(u)
dx = a·sec²(u) du

Therefore ∫1/(a² + x²) dx = ∫a·sec²(u) / a²(1 + tan²(u)) du = 1/a ∫sec²(u) / (1 + tan²(u)) du
But 1 + tan²(u) = sec²(u)
So ∫1/(a² + x²) dx = 1/a ∫ du = u/a + C

Substituting back for u (= arctan(x/a) ) gives
∫1/(a² + x²) dx = 1/a · arctan(x/a) + C

• What is the difference between x^n dx and a^x dx? That is, why is one a polynomial and one an exponential function?
• In the second function, variable x is the exponent. That is why the second one is exponential function.
• at Integrals that are inverse trigonometric functions above:

I took d/dx (arcsin x/a)=1/a *1/√1-x^2 , which does not equal 1/√a^2-x^2.

since d/dx arcsin x=1/√1-x^2, what did I do wrong?
• There's a small error you made. You were right on using the chain rule by multiplying the 1/a, but observe that you need to take the derivative of arcsin(x/a) w.r.t (x/a). You took it w.r.t x. So, you'd get 1/(√1-(x/a)^2) * 1/a. With some simplification, this turns into 1/(√a^2-x^2).
• How can we prove "Integrals that are inverse trigonometric functions"? It seems like those functions cannot be integrated with standard substitution integration methods.
• You can do it with the chain rule and some clever algebra. Say we want to know the derivative of arcsin(x/a) for some constant a (with appropriate domain restrictions).

Set y=arcsin(x/a). We're seeking y'.
Take the sine of both sides to get sin(y)=x/a.
Differentiate. We get cos(y)·y'=1/a.
Isolate y' to get y'=1/[a·cos(y)]
Use the Pythagorean identity to replace cos(y) with an expression in sin(y). We get
y'=1/[a√(1-sin²(y))]
Back-substitute sin(y)=x/a:
y'=1/[a√(1-(x/a)²)]
Now combine the fractions and simplify and you get
y'=1/√(a²-x²)

Since y=arcsin(x/a), we have that the derivative of arcsin(x/a)=1/√(a²-x²), or after integrating, that
∫1/√(a²-x²)dx=arcsin(x/a)+C.

The technique is mostly the same for the other inverse trig functions. The Pythagorean identities always relate trig functions with their derivatives, so that step always works out.