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.