Thread: Integration by Substitution - Confusion

1. Integration by Substitution - Confusion

So the integral question is:

Int(1/((1-x^2)^(-3/2)))

I thought I would substitute in x=cos(u) dx=-sin(u)du ...

Wolfram Alpha uses x=sin(u)

But ultimately, I end up with cot(u), and wolfram gets tan(u) ... Last time I checked, these two are not the same function

What went wrong?

2. Originally Posted by mike_302
So the integral question is:

Int(1/((1-x^2)^(-3/2)))

I thought I would substitute in x=cos(u) dx=-sin(u)du ...

Wolfram Alpha uses x=sin(u)

But ultimately, I end up with cot(u), and wolfram gets tan(u) ... Last time I checked, these two are not the same function

What went wrong?
change both antiderivatives back into an expression in terms of x, see what happens ...

3. Originally Posted by mike_302
So the integral question is:

Int(1/((1-x^2)^(-3/2)))

I thought I would substitute in x=cos(u) dx=-sin(u)du ...

Wolfram Alpha uses x=sin(u)

But ultimately, I end up with cot(u), and wolfram gets tan(u) ... Last time I checked, these two are not the same function

What went wrong?
You must mean $1/(1-x^2)^{3/2}= (1- x^2)^{-3/2}$ but not both "1/" and the negative exponent!

Either x= cos(u) or x= sin(u) will work because $1- cos^2(u)= sin^2(u)$ and $1- sin^2(u)= cos^2(x)$ and you get rid of that 1/2 power.

Yes, using x= cos(u) you get the integral as cot(u) and with x= sin(u) you get u= tan(x).

No, those are not the same function but since those are the results of different integrals, why would you expect them to be?

You haven't finished the problem yet!

If x= cos(u) then u= $cos^{-1}(x)$ so, with the integral equal to tan(u), you get $tan(cos^{-1}(x))$. If x= sin(u), then $x= sin^{-1}(u)$ and you get $cot(sin^{-1}(x))$.

Now, $tan= \frac{sin}{cos}$ so $tan(cos^{-1}(x))= \frac{sin(cos^{-1}(x))}{cos(cos^{-1}(x))}$. Of course, $cos(cos^{-1}(x))= x$ while $sin(cos^{-1}(x))= \sqrt{1- cos^2(cos^{-1}(x)}= \sqrt{1- x^2}$. Your integral is $\frac{\sqrt{1- x^2}}{x}$.

$cot= \frac{cos}{sin}$ so $cot(sin^{-1}(x))= \frac{cos(sin^{-1}(x))}{sin(sin^{-1}(x))}$. Now $sin(sin^{-1}(x))= x$ and $cos(sin^{-1}(x))= \sqrt{1- x^2}$ so the integral is $\frac{\sqrt{1- x^2}}{x}$, just as before.