Integration with k as constant

[ritwik06]

Homework Statement

PLease HELP ME INTEGRATE THIS>
\int\frac{dh}{\sqrt{h^{2}-k^{2}}}
k is constant

 

[Kurdt]

You will have to show some attempt first.

 

[ritwik06]

If I take h^2 common from the integral, it will be:
\int\frac{dh}{h\sqrt{1-\frac{k^{2}}{h^{2}}}}
which finally gives:
\frac{\sqrt{1-\frac{k^{2}}{h^{2}}}}{k^{2}}

 

[Defennder]

I don't see how that follows. Your first step is right, next you should use a trigo substitution.

 

[ritwik06]

I am redefining the question:
\int\frac{dx}{\sqrt{x^{2}-a^{2}}}


taking x^2 common

\int\frac{dx}{x\sqrt{1-\frac{a^{2}}{x^{2}}}}

let 1-\frac{a^{2}}{x^{2}}=t
\frac{1}{2}\int\frac{dt}{(1-t)(\sqrt{t})}
let\sqrt{t}=j
\int\frac{dj}{1-j^{2}}

how shall i proceed

 

[Kurdt]

As mentioned before trig substitution is the way to go from the very start.

 

[rootX]

I am redefining the question:
\int\frac{dx}{\sqrt{x^{2}-a^{2}}}


taking x^2 common

\int\frac{dx}{x\sqrt{1-\frac{a^{2}}{x^{2}}}}

let 1-\frac{a^{2}}{x^{2}}=t
\frac{1}{2}\int\frac{dt}{(1-t)(\sqrt{t})}
let\sqrt{t}=j
\int\frac{dj}{1-j^{2}}

how shall i proceed

I also agree with others, but assuming that you have the last step right:
\int\frac{dj}{1-j^{2}}

You should do decomposition and partial fraction ...

 

[Defennder]

You've made it a lot more tedious than it could have been with a simple trigo substitution.

 

[ritwik06]

please be more explicit. Actually there is no sign of trigo ratio in the answer :S

 

[Kurdt]

See my PDf file located in the following thread. Specifically section 4.

https://www.physicsforums.com/showthread.php?t=213784

 

[HallsofIvy]

I am redefining the question:
\int\frac{dx}{\sqrt{x^{2}-a^{2}}}


taking x^2 common

\int\frac{dx}{x\sqrt{1-\frac{a^{2}}{x^{2}}}}

let 1-\frac{a^{2}}{x^{2}}=t
\frac{1}{2}\int\frac{dt}{(1-t)(\sqrt{t})}
let\sqrt{t}=j
\int\frac{dj}{1-j^{2}}

how shall i proceed

Pretty much every step is wrong. You seem to be consistly forgetting to replace the "dx" or "dt" term.
For example, if you let 1-\frac{a^2}{x^2}=t then \frac{2a^2}{x^3}dx= dt. How are you going to substitute for that?

Even if that were correct when you say "let \sqrt{t}= j", you should have \frac{1}{2\sqrt{t}}dt= dj.

As everyone has been telling you from the start, use a trig substitution. sin^2 \theta+ cos^2 \theta= 1 so tan^2 \theta+ 1= sec^2 \theta.