What is the integral of 1-tan^2theta from 0 to pi/3?

Using the identity tan^2theta=sec^2theta-1,
I got my answer to be

2pi/3 - sqrt*3

Can someone verify this for me please?

You are correct!!

Good job :)

To find the integral of 1 - tan^2(theta) from 0 to pi/3, you can start by using the identity tan^2(theta) = sec^2(theta) - 1.

The integral becomes:

∫ (1 - tan^2(theta)) d(theta) from 0 to pi/3.

Using the identity, we get:

∫ (1 - (sec^2(theta) - 1)) d(theta) from 0 to pi/3.

Simplifying further:

∫ (2 - sec^2(theta)) d(theta) from 0 to pi/3.

Now, let's integrate term by term.

The integral of 2 d(theta) is 2(theta).

The integral of sec^2(theta) d(theta) is tan(theta).

So, the integral becomes:

2(theta) - tan(theta) from 0 to pi/3.

Now, substitute the limits of integration:

[2(pi/3) - tan(pi/3)] - [2(0) - tan(0)].

We know that tan(0) = 0, and tan(pi/3) = √3.

Substituting these values in, we get:

[2(pi/3) - √3] - [0 - 0].

Simplifying further:

2(pi/3) - √3.

Thus, your answer of 2(pi/3) - √3 is correct.