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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

## 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?(42 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(46 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.(26 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)(14 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.(26 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(13 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

□(11 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)

- 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)

- something confused me, for calculating definite integrals we must have a graph or details which we can calculate the area from them; is it right?

But for calculating indefinite integrals, using such simple rules is enough.

Am I correct?(2 votes)- For calculating definite integrals, you can use a graph, but you can also use the second part of the fundamental theorem of calculus. For indefinite integrals, the simple rules are enough for now, but it gets way more complicated later.(3 votes)