Integration by substitution u=tan(t)

[MareMaris]

Homework Statement

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

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.

 

[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 $\int_0^{\pi/2}\frac{-1}{1+x(\sin t)^2}dt$.

 

[MareMaris]

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 $\int_0^{\pi/2}\frac{-1}{1+x(\sin t)^2}dt$.

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!

 

[haruspex]

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

 

[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

 

[SammyS]

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

Out of curiosity, what is the answer?

 

[MareMaris]

Out of curiosity, what is the answer?

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

 

[haruspex]

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

I get the same.