Find definite integra lfrom -pi/6 to pi/6 of [(tanx)^3*dx]

[MarkSA]

Hello,

Find the definite integral from -pi/6 to pi/6 of: [(tanx)^3*dx]

Could you tell me if I have set this up correctly?

Let u = tanx
du = (secx)^2dx
1/(secx)^2 * du = 1dx
1/(secx)^2 * the integral from -pi/6 to pi/6 of: u^3du
1/(secx)^2 * the integral from -pi/6 to pi/6 of: (tanx)^3dx
= 1/(secx)^2 * [1/4 * (tanx)^4] and then plug in [F(pi/6) - F(-pi/6)] to get the answer.

Thanks.

 

[tkhunny]

Re: Find the definite integral

How did that secant get outside the integral? something VERY wrong going on there.

You have lots of nice tangents sitting there. Why not use a couple of them?

\(\displaystyle tan^{2}(x)\;=\;sec^{2}(x)\;-\;1\)

Once you get that done, something else could pop to mind.

 

[pka]

\(\displaystyle f(x) = \left[ {\tan (x)} \right]^3\) is an odd function.
So the integral is 0.