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Calculus 2
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 start fraction, 1, divided by, x, end fraction? Check out this video.
Want to practice integrating exponential functions and start fraction, 1, divided by, x, end fraction? Check out this exercise.
Integrals that are inverse trigonometric functions
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- 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)
- Could someone please provide me with the proof for
integral of 1/(a^2 + x^2)(3 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
□(2 votes)
- Why is the integral of tan(x) not listed?(2 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(3 votes)
- Why isn't there an arccos integral function?(2 votes)
- probably because arcsin and arccos have almost identical derivatives. their derivatives are negatives of each other.(2 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.(2 votes)
- I agree some of the things are not explained, I can't find the videos(5 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?(2 votes)
- There are proofs out there for each trig function but it is much easier to just learn them by heart.(2 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(4 votes)
- Any mnemonics to remember these? Anyone?(2 votes)
- Just commit the derivatives to memory and then use the opposites to remember these!(1 vote)
- Under the Integrals that are inverse trigonometric functions, why there is "1/a" before arctan but not arcsin?(2 votes)
- Hello,
I one of the practice problems under u substitution: definite integrals gave the problem: ∫ 1/1+9x^2 dx. If you use u substitution you get (1/3)arctan(3x)+c. But if you refer to the integral above it states that ∫1/a^2 +x^2 dx = 1/a arctan (x/a)+c. In my example a=1 but the fraction for the solution isn't 1/1 it's 1/3. I'm confused. If you could please get back to me I'd really appreciate it. Thanks.(2 votes)- When you use u sub to put u in for 3x, you are also subbing in du/3 for dx, let me know if that doesn't make sense. But it should look like this now ∫1/1+u^2 du/3 Anyway, now with du/3 or in other words (1/3)du, you can bring that 1/3 out to the front, and make it look like (1/3)∫1/1+u^2 du and solve ∫ 1/1+u^2 du = arctan(u/1)+c = arctan(3x/1)+c, but you still have that 1/3 out in front. So it's (1/3)arctan(3x)+c. Again, let me know if that doesn't make sense.(1 vote)