How do I find the primitive of $f(x)=arctanx^2+arccotanx^2$

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\tan^{-1}(x^2)+\cot^{-1}(x^2)$. I need to find the primitive function of $f$.
Now, I know that f is a constant function, because its $f'(x)=0$. And I think that let's say if $F:\mathbb{R}\rightarrow\mathbb{R}$ is its primitive then $F'(x)=f(x)$ which means that $F'(x)$ needs also to be a constant. I think I managed to solve it $\int \tan^{-1}(x^2)+\cot^{-1}(x^2)dx=\int \frac{\pi}{2}dx=\frac{\pi}{2}x+C$, am I right?

Answer 1

Yes, you are correct.

From,

$$\tan(x^2) = \operatorname{cot}\left(\frac \pi 2 - x^2\right)$$

We have,

$$\arctan(x^2) = \frac \pi 2 - \operatorname{arccot}(x^2)$$

Hence,

$$\int (\arctan(x^2) + \operatorname{arccot}(x^2)) \mathrm dx = \int \left(\frac \pi 2 - \operatorname{arccot}(x^2) + \operatorname{arccot}(x^2)\right)\mathrm dx $$

Which is quite easy to evaluate,

$$\int \frac \pi 2 \mathrm dx = \frac \pi 2 x + C$$

Proof of first line

$$\operatorname{cot}\left(\frac \pi 2 - x^2\right) = \frac{\cos\left(\frac \pi 2 - x^2\right)}{\sin\left(\frac \pi 2 - x^2\right)} = \frac{\sin(x^2)}{\cos(x^2)} = \tan(x^2)$$