What is cosine to the power of zero?

I was doing a question relates to substitution rule under integration.

The question is as follow:

Evaluate $\int{1\over{(1+x^2)^n}}dx, n\in \mathbb{Z}^+$

We have seen that $(\tan^{-1}x)'=\frac{1}{1+x^2}$

Let $x= \tan t, t\in(-\frac{\pi}{2},\frac{\pi}{2})$. Then $\frac{dx}{dt}=\sec^2t$

$$\int\frac{1}{(1+x^2)^n}dx=\int\frac{1}{(1+tan^2t)^n}\sec^2tdt$$ $$=\int\frac{\sec^2t}{\sec^{2n}t}dt=\int\cos^{2n-2}tdt$$

so I got the final answer $\cos^{2n -2} (t) dt$, but the question continues to ask what's the integral when $n=1$.

I substitute $n=1$, and I got $\cos^0 (t)$ which is totally unfamiliar to me.

So what is $\cos^0 (t)$ ?

Solution 1:

$\cos^0(t) = 1$ if $\cos t \ne 0$ and undefined else, so $\int \cos^0 t \ \mathrm dt = \int 1 \ \mathrm dt = t + C$. This results in $$\int \frac1{(1+x^2)^1} \ \mathrm dx = t + C= \arctan x + C$$ wich is consistent with what you know. Note that $\arctan$ is a nicer way of writing $\tan^{-1}$.

Solution 2:

For every non-zero number $x$, you have $x^0=1$