Answer: For \(n \geqslant 2\),

\(I_n=\int_0^{\pi/4} \tan^n x\,\mathrm{d}x=\int_0^{\pi/4} \tan^{n-2}x\,\tan^2x\,\mathrm{d}x\).

Using \(\tan^2x=\sec^2x-1\),

\(I_n=\int_0^{\pi/4} \tan^{n-2}x\,(\sec^2x-1)\,\mathrm{d}x\)

\(=\int_0^{\pi/4} \tan^{n-2}x\sec^2x\,\mathrm{d}x-I_{n-2}\).

Now let \(u=\tan x\), so \(\mathrm{d}u=\sec^2x\,\mathrm{d}x\). Then

\(\int_0^{\pi/4} \tan^{n-2}x\sec^2x\,\mathrm{d}x=\int_0^1 u^{n-2}\,\mathrm{d}u=\left[\frac{u^{n-1}}{n-1}\right]_0^1=\frac{1}{n-1}.\)

Hence \(I_n=\frac{1}{n-1}-I_{n-2}\).

Also \(I_0=\int_0^{\pi/4}1\,\mathrm{d}x=\frac{\pi}{4}\).

So

\(I_2=\frac{1}{1}-I_0=1-\frac{\pi}{4}\),

\(I_4=\frac{1}{3}-I_2=\frac{1}{3}-1+\frac{\pi}{4}\),

\(I_6=\frac{1}{5}-I_4=\frac{1}{5}-\frac{1}{3}+1-\frac{\pi}{4}\),

and

\(I_8=\frac{1}{7}-I_6=\frac{1}{7}-\frac{1}{5}+\frac{1}{3}-1+\frac{\pi}{4}.\)

(a) For \(n\geqslant 2\), write

\(I_n=\int_0^{\pi/4}\tan^n x\,\mathrm{d}x=\int_0^{\pi/4}\tan^{n-2}x\,\tan^2x\,\mathrm{d}x.\)

Since \(\tan^2x=\sec^2x-1\),

\(I_n=\int_0^{\pi/4}\tan^{n-2}x(\sec^2x-1)\,\mathrm{d}x\)

\(=\int_0^{\pi/4}\tan^{n-2}x\sec^2x\,\mathrm{d}x-I_{n-2}.\)

Now use \(u=\tan x\), so \(\mathrm{d}u=\sec^2x\,\mathrm{d}x\). When \(x=0\), \(u=0\); when \(x=\pi/4\), \(u=1\). Therefore

\(\int_0^{\pi/4}\tan^{n-2}x\sec^2x\,\mathrm{d}x=\int_0^1 u^{n-2}\,\mathrm{d}u=\left[\frac{u^{n-1}}{n-1}\right]_0^1=\frac{1}{n-1}.\)

So the reduction formula is

\(I_n=\frac{1}{n-1}-I_{n-2}.\)

(b) First evaluate

\(I_0=\int_0^{\pi/4}1\,\mathrm{d}x=\frac{\pi}{4}.\)

Then apply the recurrence repeatedly:

\(I_2=\frac{1}{1}-I_0=1-\frac{\pi}{4},\)

\(I_4=\frac{1}{3}-I_2=\frac{1}{3}-1+\frac{\pi}{4},\)

\(I_6=\frac{1}{5}-I_4=\frac{1}{5}-\frac{1}{3}+1-\frac{\pi}{4},\)

\(I_8=\frac{1}{7}-I_6=\frac{1}{7}-\frac{1}{5}+\frac{1}{3}-1+\frac{\pi}{4}.\)