Reduction Formulae of $I_n=\int_0^\frac{\pi}{4}\tan^nx\, dx$ and $J_n=(-1)^nI_{2n}$

Question

Let $I_n=\int_0^\frac{\pi}{4}\tan^nx\,dx$ and let $J_n=(-1)^nI_{2n}$ for $n=0,1,2$

1. Show that $I_n+I_{n+2}=\frac{1}{n+1}$.
2. Deduce that $J_n-J_{n-1}=\frac{(-1)^n}{2n-1}$ for $n\ge1$
3. Show that $J_m=\frac{\pi}4+\sum_{n=1}^{m}\frac{(-1)^n}{2n-1}$.
4. Use the substitution $u=\tan x$ to show that $I_n=\int_0^1\frac{u^n}{1+u^2}du$.
5. Deduce that $0 \le I_n \le \frac{1}{n+1}$ and conclude that $J_n \to 0$ as $n \to \infty$.

This is for the Proofs topic of the Extension 2 HSC maths course. This belongs to a section using recurrence integrals, but I think there is a bit of inequalities proof at the end.

I have attempted the first section. I could proof that $I_n+I_{n-2}=\frac1{n-1}$ by splitting $\tan^nx$ into $\tan^{n-2}x \cdot \tan^2x$. I can see that it is in some ways similar, but I can't quite get it.

I have successfully done part four:

$I_n=\int_{0}^{\frac{\pi}4}\tan^nx\,dx$

$u=\tan x$; $\frac{du}{dx}=\sec^2x$

$I_n=\int_0^1u^n\frac{dx}{\sec^2x}=\int_0^1\frac{u^n}{1+u^2}du$

I know this is a really long question, but I really don't understand how to do it. Any help would be really appreciated. Thank you!!!