Question

Help simply this integral?

Answer 1

\[\large \int\limits_{}^{}\frac{ 1 }{ x^2+a^2 }~dx\]

Answer 2

I know it has something to do with the inverse tan

Answer 3

It is

Answer 4

help me microwave

Answer 5

shut up now i wont

Answer 6

??? :(

Answer 7

stop playing games kiddo

Answer 8

lol I think you're the one playing games with me

Answer 9

my guess would be treat a as number

Answer 10

try this example

http://www.pleacher.com/handley/gifs5/trigsub6.gif

Answer 11

That only shows the answer :(

Answer 12

Recall your Pythagorean Identity:
\(\large\rm 1+\tan^2\theta=\sec^2\theta\)
If we multiply through by a^2
\(\large\rm a^2+a^2\tan^2\theta=a^2\sec^2\theta\)
which we can write this way,
\(\large\rm a^2+(\color{indianred}{a\tan\theta})^2=a^2\sec^2\theta\)

So if we look back our problem,\[\large\rm \int\limits\frac1{a^2+(\color{indianred}{x})^2}~dx\]We can see that if we replaced x with a*tan(theta),
we would be able to `eliminate the addition` in the denominator by applying our Pythagorean Identity! That's the point of process, trig identities get rid of these problematic operations like addition and subtraction.

Answer 13

So making your substitution,\[\large\rm \int\limits\limits\frac1{a^2+(\color{indianred}{a \tan \theta})^2}~dx\]We see that our denominator simplifies nicely,\[\large\rm \int\limits\limits\limits\frac1{a^2 \sec^2\theta}~dx\]We still need to do some work replacing our differential dx with something involving d(theta) though.

Answer 14

\[\large\rm x=a \tan \theta\qquad\to\qquad dx=a \sec^2\theta~d \theta\]

Answer 15

\[\large\rm \int\limits\limits\limits\limits\frac1{a^2 \sec^2\theta}~a \sec^2\theta~d \theta\qquad=\qquad \frac1 a\int\limits d \theta\]So it becomes significantly easier to integrate, ya?

Answer 16

\[\large\rm =\frac1a \theta+c\]Then rewriting this in terms of x, and not theta, requires just a couple more steps.

Answer 17

Aren't you a gem oh

Answer 18

lol microwave :D must be some inside joke that i dont get

Answer 19

Some ppl yknow zeppy are just... ruthless. Ugly piece of steamed rice is what I like to call them. :D

Answer 20

Steve you deserved that one

Answer 21

lol XD

Answer 22

zepdrix you are amazing

Answer 23

You're welcome you ugly piece of steamed rice :D
lol

Answer 24

So basically\[\large x=a \tan \theta\]\[\large\frac{ x }{ a }=\tan \theta\]\[\large \theta = \tan^{-1}\frac{ x }{ a }\]

Answer 25

Amirite?

Answer 26

\[\large\rm =\frac1a \theta+c\]Ya that looks right :)\[\large\rm =\frac1a \arctan\frac x a+c\]

Answer 27

aha thank you lol

Answer 28

"You're welcome you ugly piece of steamed rice"

I agree with Steve, Zepdrix you're amazing.

Medaled

Answer 29

lol

Answer 30

ShadowLegendX Best Response Medals 0
"You're welcome you ugly piece of steamed rice"

I agree with Steve, Zepdrix you're amazing.