Question

Simple arctan integration

Answer 1

\[\int\limits_{?}^{?} \frac{ 1 }{ 4+x^2 } dx\]

Answer 2

I multiplied top and bottom by 1/4, i get \[\frac{ \frac{ 1 }{ 4 } }{ 1 + \frac{ x^2 }{ 4 } }\]

Answer 3

Take out the 1/4 ( and do arctan(x/2)

Answer 4

What did i do wrong?

Answer 5

\[\int\limits_{}^{}\frac{ 1 }{ x^2+a^2 }~dx=\frac{1}{a}\arctan(\frac{x}{a})\]

Answer 6

\[\int\limits_{}^{}\frac{1}{x^2+4}~dx=\int\limits_{}^{}\frac{1}{x^2+2^2}~dx=\frac{1}{2}\arctan(\frac{x}{2})+c\]

4 became z^2 and that 2 replaced with a in a^2

Answer 7

4 became 2^2 and that 2 replaced with a in a^2 *

Answer 8

makes sense?

Answer 9

I thought the derivative of tan^-1 is 1/1+x^2

Answer 10

Yes, 1+x^2 can work, i rewrite it as x^2+1 but that doesnt change, i can both apply to the rule either or. Sorry, i didn't mean to confuse by rearranging it to make the higher degree bethe first.
you can still apply this \[\int\limits_{}^{}\frac{1}{a^2+x^2}~dx=\int\limits_{}^{}\frac{1}{2^2+x^2}~dx=\frac{1}{2}\arctan(\frac{x}{2})+c\]

Answer 11

it can both*

Answer 12

i don't really see why u can do 1/a arctan (x/a) , maybe can u tell me what is wrong with my?

Answer 13

I just saw the proof of 1/a arctan(x/a) , it was interesting

Answer 14

did you apply the u-sub after you factored out 1/4

Answer 15

i see now, you have to do u -sub after that lol.

Answer 16

thank you

Answer 17

No problem

Answer 18

Do you mind answering 1 more question?

Answer 19

\[\int\limits_{-4}^{4}xcos(x)dx\] why does this equal to 0??

Answer 20

The rule i gave you, is a good shortcut of these kind of problem before actually getting to do the u-sub part. Because you wont spend your time doing this when you get to learn other techniques with integrations.

Answer 21

yeah i understand what u mean, i will try to use that rule when i see it

Answer 22

\(STEP~1:\)do integration by parts first, to take the anti derivative
\(STEP~2:\) and use the fundemental theorem of calculus. \(\int\limits_{a}^{b} f(x)=F(b)=F(a)\)

Answer 23

You apply integration by parts as soon as you see not parent functions multiplying together and if you can't u-sub, then you might as well try integration by parts
Integration by parts is:
\[\int\limits_{}^{} udv=u*v - \int\limits_{}^{} vdu\]
Pick a u, then take the derivative for du
Pick a dv, then integrate it for V