# Derivation of some integral formulas

1. Mar 14, 2015

### Edd257

Hi,

The integrals in the image are well-known, but we are not allowed to use them in the test.

I have tried to prove them using changing variable t, but I didn't succeed.

Can you show me the way?

2. Mar 14, 2015

### Ray Vickson

Show us your actual work on at least one example above---whether it failed or not.

3. Mar 14, 2015

### HallsofIvy

Staff Emeritus
What do you mean by "not allowed to use them"? How would being able to prove them help you on a test?

The first and fourth can be done by considering the derivatives of arctan and arcsin. The second and third by factoring and "partial fractions".

4. Mar 14, 2015

### SammyS

Staff Emeritus
For the first integral, try the trig substitution, $x = a\cdot\tan(\theta)\$.

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5. Mar 14, 2015

### fourier jr

Geez I didn't think it was necessary to use a trig substitution for any of these. In the first three you can just multiply the numerator & denominator by 1/a2 & in the last two multiply numerator & denominator by 1/a.

6. Mar 14, 2015

### Edd257

ok,

I succeed proving number one and four- please check if I did it correctly.
But I still didn't understand how to prove the rest... how do I use partial fractions?
can you show me the way on one of the other integrals?

7. Mar 15, 2015

### Edd257

I have tried to do that in your way in the second integral and didn't succeed.
Can you show me your way?

8. Mar 16, 2015

### torstein

Hey, I'm on my mobile right now so I can't see your proves for first and four integrals but for the second and third use simple fractions. I mean, you have the denominator

a^2 - x^2 = (a+x)(a-x)

So you can write

1/(a^2 - x^2) = 1/((a+x)(a-x)) = A/(a+x) + B/(a-x)

For some constants A, B which you should find easily then you just have a sum of two integrals which are two logarithms and using a basic property of log you will get what you are looking for. Third integral is analogous to this.

Should I said why take abs value?

9. Mar 17, 2015

### Edd257

Tnx, now I got it.

But I still didn't understand how to solve the last integral...

I need to use some substitution t?

10. Mar 17, 2015

### fourier jr

I think in the last one you need to use the substitution x = a⋅sec θ then dx = a⋅secθ⋅tanθ dθ etc

11. Mar 17, 2015

### SammyS

Staff Emeritus
That works for the " - " sign. Alternatively, you can use x = a ⋅ cosh(u) .

Use x = a⋅sinh(u) for the "+" sign.

Of course the results will look a bit different: cosh-1(x/a) & sinh-1(x/a) .