# Math Help - Relating a function and its derivative?

1. ## Relating a function and its derivative?

I am trying to find the definite integral of tanx^3, with pi/6 at the top of the integral sign and -pi/6 on the bottom.

The only problem I have is that I do not see how this fits the profile of:

f(g(x))*g'(x)

What would the g(x) be in tanx^3? Once I have that I have no problems finding the rest.

2. Originally Posted by redman223
I am trying to find the definite integral of tanx^3, with pi/6 at the top of the integral sign and -pi/6 on the bottom.

The only problem I have is that I do not see how this fits the profile of:

f(g(x))*g'(x)

What would the g(x) be in tanx^3? Once I have that I have no problems finding the rest.
$\int\tan^3(x)dx=\int\tan(x)(\sec^2(x)-1)dx=$ $\int\bigg[\tan(x)\sec^2(x)-\tan(x)\bigg]dx$

The last one you should know how to do.

If not, for the first part let $u=\tan(x)$ and for the second rewrite it as a quotient of sine and cosine and then let $u=\cos(x)$

3. I don't understand what to do with u=tanx

If I use tanx for u, then u' would be secx^2, but what would f(x) be?

I only know how to solve these with the equation f(g(x))*g'(x), where g(x) is u, but I am having trouble setting it up.

4. Originally Posted by redman223
I don't understand what to do with u=tanx

If I use tanx for u, then u' would be secx^2, but what would f(x) be?

I only know how to solve these with the equation f(g(x))*g'(x), where g(x) is u, but I am having trouble setting it up.
$u=\tan(x)\Rightarrow{du=\sec^2(x)}$

Or you can see that $f(x)=\left(\tan(x)\right)^1$ and $g(x)=\tan(x)\Rightarrow{g'(x)=\sec^2(x)}$

5. So tanx*secx^2 = tanx^3?

If that is true, I am having trouble seeing it.

6. Originally Posted by redman223
So tanx*secx^2 = tanx^3?

If that is true, I am having trouble seeing it.
No... $\tan(x)\times\left[\sec^2(x)-1\right]=tan^3(x)$

But we need to distribute it through...

thus, we get $\tan(x)\sec^2(x)-\tan(x)$

Integrating we have $\int \tan(x)\sec^2(x)\,dx-\int\tan(x)\,dx$

Focusing on the first integral...
$\int \tan(x)\sec^2(x)\,dx$

Let $u=\tan(x)$ [note that the value for u is also our g(x)]. Thus, $\frac{du}{dx}=\sec^2(x)$ [again, this is the same as g'(x)].

Thus, what we are left with then is $\int u\,du=\frac{1}{2}u^2=\frac{1}{2}\tan^2(x)$

Now the second integral...

$\int\tan(x)\,dx$

Rewriting the integrand you have $\int\frac{\sin(x)}{\cos(x)}\,dx$

Let $u=\cos(x)$ [this is our g(x)]. Thus, $\frac{du}{dx}=-\sin(x)$ [this is g'(x)].

However, this doesn't quite match the g'(x) value in the integrand $\sin(x)$. To fix this, multiply by $\frac{-1}{-1}$

Thus, we see the integral become $-\int\frac{\,du}{u}=-\ln|u|=-\ln|\cos(x)|$.

Combining the two, we see $\int\tan^3(x)\,dx=\int\left[\tan(x)\sec^2(x)-\tan(x)\right]\,dx=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|$

Now evaluate from $-\frac{\pi}{6}$ to $\frac{\pi}{6}$

Does this make a little more sense?

7. The only thing that I am having trouble understanding is why tan(x)sec(x)^2-tan(x) = tan(x)^3.

8. Hey you dont need all that!

$I = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \tan^3 x \, dx$

Substitute $x \to -x, dx \to -dx$

$I = \int_{\frac{\pi}{6}}^{-\frac{\pi}{6}} \tan^3 (-x) \, (-dx)$

$I = -\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} -\tan^3 (x) \, (-dx)$

$I = -\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \tan^3 (x) \, dx$

$I = -I \Rightarrow 2I = 0 \Rightarrow I = 0$

In fact for any odd function f(x), and any real number a,

$\int_{-a}^a f(x)\, dx = 0$