Question

i cant figure out the result of this integral using two methods:

int ((a^2)+(x^2))^(-3/2)dx

i used one method of substituting x with a*tan(theta).
can anyone figure out another method? simpler is better.

thanks

Answer 1

i cant think of it... doing the trig sub seems simple enough to me

Answer 2

use this substitution
u=a * tan x
then a^2+x^2=a^2(1+tan^2x )=a^2 sec^x

Answer 3

@amoodarya , read the whole post, that trig sub is already been mentioned

Answer 4

its pretty easy
Substitute
x=\[a \tan \theta\]
then dx=\[a \sec ^2\theta d \theta\]

If the integral is I then now
I= \[\int\limits_{}^{} \frac{ a \sec ^2\theta d \theta }{ (asec \theta)^3}\]

I=\[\int\limits_{}^{}\frac{ 1 }{ a^2 }* \cos \theta d \theta\]

I=\[\frac{ 1 }{ a^2 } \sin \theta + c\]

I= \[\frac{ 1 }{ a^2} * \frac{ x }{ \sqrt{(a^2+x^2)} } + c\]

Answer 5

Oh you asked for another method :P I didn't read the entire question. You can do this with integration by parts but trigonometric substitution is the shortest and simplest

Answer 6

you can use also u=sinh x
so
a^2+x^2=a^2(1+sinh^2x)=a^2 cosh^2x
and
du= cosh x dx

Answer 7

thanks MayankD for the reply.

How can I compute it by parts?

Answer 8

Use \[\int\limits_{}^{}u*v*dx=u \int\limits_{}^{}v*dx - \int\limits_{}^{}[u' *\int\limits_{}^{}v*dx]dx\]

Answer 9

how can I split the function to: \[\int\limits_{}^{}u*v*dx\]
thankyou for the reply

Answer 10

Take u as 1 and v as the rest of the function. Its a difficult integral to solve but with substitution you should be able to do it sooner or later.

Answer 11

thankyou MayankD

yes I agree that with substitution it is very simplier to solve the integral.