Integrating (tan(x/2))^2 between 0 and pi

[Jbreezy]

Integral of ...

Homework Statement

Hi, no directions were given it just says ∫(tan(x/2))^2 dx between 0 and pi.
You will get for the integral (1/2 (sin(2x)) - ((1/6)sin(2x))^3
I think that this is OK. Part of the graph of the origonal function dips below the axis so it end up being 0. I should change the limits of integral between [0,pi/2] then multiply that answer by 4 because there are 4 areas of that size.
Is this right?
So my final answer I got (4 (1/3)) so 4/3.


Thanks

Homework Equations

The Attempt at a Solution

 

[CAF123]

Homework Statement

Hi, no directions were given it just says ∫(tan(x/2))^2 dx between 0 and pi.
You will get for the integral (1/2 (sin(2x)) - ((1/6)sin(2x))^3

How did you get that expression? Sketch the graph of tan(x/2) and observe the behaviour at x = π. What does this suggest about the integral?

 

[Jbreezy]

I wrote the wrong integral. I mean integral of (cos(2x))^3 between 0 and Pi.
Now read the rest of what I wrote. I'm so sorry.

 

[CAF123]

I wrote the wrong integral. I mean integral of (cos(2x))^3 between 0 and Pi.
Now read the rest of what I wrote. I'm so sorry.

If the question is just ##\int_0^\pi (\cos(2x))^3 dx##, then as you said the answer is zero.
You computed the integral of the modulus of the integrand over the same interval of integration.

If the question asked something like 'Find the area enclosed by cos^3(2x)..'', for example, then do as you did.

 

[Jbreezy]

It doesn't say anything it just has the integral. So I guess I will go with 0 then. Thanks dude. Sorry about the mistake.