# Integration with k as constant

1. Aug 5, 2008

### ritwik06

1. The problem statement, all variables and given/known data

$$\int\frac{dh}{\sqrt{h^{2}-k^{2}}}$$
k is constant

2. Aug 5, 2008

### Kurdt

Staff Emeritus
Re: Integration:

You will have to show some attempt first.

3. Aug 5, 2008

### ritwik06

Re: Integration:

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:
$$\frac{\sqrt{1-\frac{k^{2}}{h^{2}}}}{k^{2}}$$

4. Aug 5, 2008

### Defennder

Re: Integration:

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

5. Aug 9, 2008

### ritwik06

Re: Integration:

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

6. Aug 9, 2008

### Kurdt

Staff Emeritus
Re: Integration:

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

7. Aug 9, 2008

### rootX

Re: Integration:

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 ...

8. Aug 9, 2008

### Defennder

Re: Integration:

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

9. Aug 9, 2008

### ritwik06

Re: Integration:

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

10. Aug 9, 2008

### Kurdt

Staff Emeritus
11. Aug 9, 2008

### HallsofIvy

Staff Emeritus
Re: Integration:

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 $$tan^2 \theta+ 1= sec^2 \theta$$.