Question

integral of (x^2-a^2)^(3/2) dx ... ?

Answer 1

\[\int\limits_{}^{}(x^2-a^2)^\frac{3}{2} dx\]
TRig substition! :)

Answer 2

a nice trig sub for myininaya, queen of the trig substitutions

Answer 3

yes but how to use it ? i don't see it

Answer 4

actually i think this is kind of a pain

\[x=a\tan(\theta)\] or
\[x=a\sec(\theta)\] probably the second one

Answer 5

It's the second one.

Answer 6

x=a*cosh(u) works also

Answer 7

\[x=a\sec(\theta),dx=a\sec(\theta) \tan(\theta)d\theta\] then
\[\sqrt{a^2\sec^2(\theta)-a^2}=a\tan(\theta)\] so you get
\[\int a^3\tan^3(\theta) \sec(\theta)\tan(\theta)d\theta\]

Answer 8

so we will use a sec sub.
you coud use csc if you like though (rearrage your triangle)

\[otherleg=\sqrt{x^2-a^2}\]

\[\sec(\theta)=\frac{x}{a}\]
\[\tan(\theta) \sec(theta) d\theta=\frac{1}{a} dx\]
\[\int\limits_{}^{}((a sec(\theta))^2-a^2)^\frac{3}{2} a \sec(\theta) \tan(\theta) d \theta\]
\[\int\limits_{}^{} (a^2)^\frac{3}{2}(\tan^2(\theta))^\frac{3}{2} \sec(\theta) \tan(\theta) d \theta\]

Answer 9

i knew you had it!

Answer 10

\[\int\limits_{}^{} a^3 \tan^3(\theta) \sec(\theta) \tan(\theta) d \theta \]

Answer 11

oh you got what i got. now you still have to integrate
\[a^3\int\tan^4(\theta)\sec(\theta)d\theta\]

Answer 12

\[\int\limits_{}^{} a^3 \tan^4(\theta) \sec(\theta) d \theta \]

\[ a^3 \int\limits_{}^{} (\sec^2(\theta)-1)^2 \sec(\theta) d \theta \]
\[a ^3 \int\limits_{}^{}(\sec^4(\theta)-2 \sec^2(\theta) +1) \sec(\theta) d \theta\]
\[a^3 \int\limits_{}^{}( \sec^5(\theta) -2 \sec^3(\theta)+\sec(\theta) ) d \theta\]

Answer 13

so looks at
\[\int\limits_{}^{}\sec(\theta) \cdot \frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)} d \theta=\int\limits_{}^{} \frac{du}{u}=\ln|u|+c_1\]
\[=\ln|\sec(\theta)+\tan(\theta)|+c_1\]
now we will look at
\[\int\limits_{}^{}\sec^3(\theta) d \theta=\int\limits_{}^{}\sec^2(\theta) \sec(\theta) d \theta\]
\[\int\limits_{}^{}\sec^3(\theta) d \theta=\tan(\theta) \sec(\theta)-\int\limits_{}^{}\tan(\theta) \tan(\theta) \sec(\theta) d \theta\]
\[\int\limits_{}^{}\sec^3(\theta) d \theta=\tan(\theta) \sec(\theta)-\int\limits_{}^{}(\sec^2(\theta)-1)\sec(\theta) d \theta\]
\[\int\limits_{}^{}\sec^3(\theta) d \theta=\tan(\theta) \sec(\theta) -\int\limits_{}^{}\sec^3(\theta) d \theta +\int\limits_{}^{}\sec(\theta) d \theta\]
now add that integral of sec^3(theta) d theta on both sides
\[2 \int\limits_{}^{}\sec^3(\theta) d \theta=\tan(\theta) \sec(\theta) +\ln|\sec(\theta)+\tan(\theta)|+c_2\]

Answer 14

now that last part
\[\int\limits_{}^{}\sec^5(\theta) d \theta=\int\limits_{}^{} \sec^3(\theta) \sec^2 (\theta)d \theta\]
\[\int\limits_{}^{} \sec^3(\theta) (1+\tan^2(\theta)) d \theta\]
\[\int\limits_{}^{}\sec^3(\theta) d \theta +\int\limits_{}^{}\sec^3(\theta) \tan^2(\theta) d \theta\]
\[\frac{1}{2}(\tan(\theta) \sec(\theta)+\ln|\sec(\theta)+\tan(\theta)|)+\int\limits_{}^{}\sec^3(\theta) \tan^2(\theta) d \theta\]

Answer 15

but you have x not theta...

Answer 16

just went for a sandwich and a beer, came back and you are still hard at work...

Answer 17

we still aren't finishing integrating yet

Answer 18

i can only do one thing at a time

Answer 19

you finish it i dare you

Answer 20

i don't know how

Answer 21

i need a break
its long

Answer 22

some people like calc two...

Answer 23

this is why god invented maple

Answer 24

oh and wolfram. but that gets it

Answer 25

ok well i should have written
=\[\int\limits_{}^{}\sec^5(\theta) d \theta=\int\limits_{}^{}\sec^4(\theta) \sec(\theta) d \theta\]
=\[\int\limits_{}^{}(\tan^2(\theta)+1)^2 \sec(\theta) d \theta \]
=\[\int\limits_{}^{}(1+2 \tan^2(\theta) +\tan^4(\theta)) \sec(\theta) d \theta \]
=\[\int\limits_{}^{} \sec(\theta) d \theta +\int\limits_{}^{} 2 \tan^2( \theta) \sec(\theta) d \theta+ \int\limits_{}^{} \tan^4(\theta) \sec(\theta) d \theta\]
=\[\ln|\sec(\theta)+\tan(\theta)| +2 \int\limits_{}^{} (\sec^2(\theta)-1) \sec(\theta) d \theta + \int\limits_{}^{} (\sec^2(\theta)-1)^2 \sec(\theta) d \theta \]

Answer 26

http://www.wolframalpha.com/input/?i=integrate+%28x%5E2+-+a%5E2%29%5E%283%2F2%29