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### Course: AP®︎/College Calculus BC > 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

## Radicals

*Want to learn more about integrating polynomials and radicals? Check out this video.*

*Want to practice integrating polynomials and radicals? Check out these exercises:*

## Trigonometric functions

*Want to learn more about integrating trigonometric functions? Check out this video.*

*Want to practice integrating trigonometric functions? Check out these exercises:*

## Exponential functions

## Integrals that are logarithmic functions

*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

## Want to join the conversation?

- Why isn't there an arccos integral function?(46 votes)
- 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(53 votes)

- 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.(28 votes)
- I think this one is the one you are looking for:

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-1b/v/exponential-functions-differentiation-intro

(I know this is a bit late, but perhaps other people can get use out of this)(16 votes)

- 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?(8 votes)
- There are proofs out there for each trig function but it is much easier to just learn them by heart.(27 votes)

- Why is the integral of tan(x) not listed?(8 votes)
- If you just want to know the answer, then Sal covers it in this video:

https://www.khanacademy.org/math/calculus-home/integration-techniques-calc/reverse-chain-rule-calc/v/integral-of-tan-x

If you feel it's an omission that needs correction (I'd agree) then you might like to raise a "feature request":

https://khanacademy.zendesk.com/hc/en-us/community/topics/200136634-Feature-Requests(14 votes)

- Any mnemonics to remember these? Anyone?(8 votes)
- Just commit the derivatives to memory and then use the opposites to remember these!(5 votes)

- Could someone please provide me with the proof for

integral of 1/(a^2 + x^2)(4 votes)- 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

□(12 votes)

- What is the difference between x^n dx and a^x dx? That is, why is one a polynomial and one an exponential function?(3 votes)
- In the second function, variable x is the exponent. That is why the second one is exponential function.(6 votes)

- 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?(2 votes)- 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).(7 votes)

- How can we prove "Integrals that are inverse trigonometric functions"? It seems like those functions cannot be integrated with standard substitution integration methods.(3 votes)
- 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.(5 votes)

- in common integral review under exponeential functions how integration of ax is ax/ln{a} shoudnt it be like the polynomial example?(1 vote)
- a^x is not a polynomial(5 votes)