# Integration by substitution u=tan(t)

1. Nov 28, 2012

### MareMaris

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

Integrate -1/(1+x(sin(t))^2) between 0 and pi/2 using the substitution u = tan(t)

3. The attempt at a solution

du/dt = (sec(t))^2
dt/du = 1/(1+u^2)

I've messed around with the integral and trig. identities but I don't seem to be getting anywhere changing the integral to make it easier to integrate.

2. Nov 28, 2012

### Michael Redei

No idea if this would help, but I'd try writing dt/du = 1/(1+u²) as dt = du/(1+u²) and substitute that in the integral $\int_0^{\pi/2}\frac{-1}{1+x(\sin t)^2}dt$.

3. Nov 28, 2012

### MareMaris

Hmm, I had a go at that but I'm still left with this sin^2(t) which I'm not sure how to get rid of, I tried changing it to 1-(cos^t) and tried linking that to dx/du which is cos^2(t) but no joy so far!

4. Nov 28, 2012

### haruspex

sin2 = 1 - cos2
cos2 = sec-2
sec2 = 1 + tan2

5. Nov 28, 2012

### MareMaris

In the end to get sin(t) and cos(t) in terms of u I just drew a right angle triangle with the side opposite to the hypotenuse as u and the side adjacent to it as 1, and worked it out using trig, seemed easier than using the identities and double angle rules that potentially you have to use because I know you use those to work out sin(t) and cos(t) with you use t sub.

I managed to get to an answer using sin(t) = u/(1+u^2)^1/2

6. Nov 28, 2012

### SammyS

Staff Emeritus
Out of curiosity, what is the answer?

7. Nov 29, 2012

### MareMaris

In the end after substituting everything in I got -pi/2root(1+x)

8. Nov 29, 2012

### haruspex

I get the same.