Evaluate the definite integral. (The answer should be 0.)?

Question

The integral (with upper limit pi/6 and lower limit -pi/6) of (tanx)^3 dx.
Please show your work. Thank you. :)

Answer 1

tan(x) is an odd function....so the integral from -a to a of (tan(x)) is always 0. Similarly, tan(x)^3 is also odd, so the integral from -pi/6 to pi/6 is zero.

Steve

PS - the actual integral is ln(cos(x))+sec(x)^2 / 2

Answer 2

First, let's solve for the integral itself and then worry about the bounds of integration later.

To solve for the integral of (tan(x))^3, split it up into a product of (tan(x))^2 and tan(x). Therefore, you now have

tan(x) * (tan(x))^2

Then, use the identity [tan(x)]^2 = (sec(x))^2 - 1,

tan(x) ([sec(x)]^2 - 1)

Distribute

tan(x) [sec(x)]^2 - tan(x)

Now, we take the integral of that. To save time, I'll split it up into two separate integrals immediately.

Integral (tan(x)[sec(x)]^2)dx - Integral (tan(x))dx

To solve for the first integral, use substitution (let u = tan(x), du = [sec(x)]^2 dx), and for the second integral convert tan(x) into sin(x)/cos(x) and use substitution (let u = cos(x), du = -sinx dx, -du = sinx dx.

You'll end up with

1/2 [tan(x)]^2 + ln|sec(x)|, which you then evaluate from -pi/6 to pi/6.