Answer: The reduction result is

\(I_n-I_{n+1}=\frac{2^{-\left(n+\frac12\right)}}{2n+1}.\)

Also,

\(\int_0^{\frac14\pi}\frac{\sin^6x}{\cos x}\,dx=\ln(1+\sqrt2)-\frac{73\sqrt2}{120}.\)

Let \(I_n=\int_0^{\pi/4} \frac{\sin^{2n}x}{\cos x}\,dx\).

Then

\(I_{n+1}=\int_0^{\pi/4} \frac{\sin^{2n+2}x}{\cos x}\,dx\).

Since \(\sin^2 x=1-\cos^2 x\),

\(I_{n+1}=\int_0^{\pi/4} \frac{(1-\cos^2 x)\sin^{2n}x}{\cos x}\,dx\)

\(=\int_0^{\pi/4} \frac{\sin^{2n}x}{\cos x}\,dx-\int_0^{\pi/4} \cos x\,\sin^{2n}x\,dx\)

\(=I_n-\int_0^{\pi/4} \cos x\,\sin^{2n}x\,dx.\)

Now evaluate the remaining integral with \(u=\sin x\), so \(du=\cos x\,dx\):

\(\int_0^{\pi/4} \cos x\,\sin^{2n}x\,dx=\left[\frac{\sin^{2n+1}x}{2n+1}\right]_0^{\pi/4}.\)

Because \(\sin(\pi/4)=\frac1{\sqrt2}\), this gives

\(\int_0^{\pi/4} \cos x\,\sin^{2n}x\,dx=\frac{(1/\sqrt2)^{2n+1}}{2n+1}=\frac{2^{-(n+\frac12)}}{2n+1}.\)

Hence

\(I_n-I_{n+1}=\frac{2^{-(n+\frac12)}}{2n+1}.\)

For the second part, \(\int_0^{\pi/4}\frac{\sin^6 x}{\cos x}\,dx=I_3\).

First,

\(I_0=\int_0^{\pi/4}\sec x\,dx=[\ln|\sec x+\tan x|]_0^{\pi/4}=\ln(1+\sqrt2).\)

Using the recurrence:

\(I_1=I_0-\frac{2^{-1/2}}{1}=\ln(1+\sqrt2)-\frac1{\sqrt2},\)

\(I_2=I_1-\frac{2^{-3/2}}{3}=\ln(1+\sqrt2)-\frac1{\sqrt2}-\frac1{6\sqrt2},\)

\(I_3=I_2-\frac{2^{-5/2}}{5}=\ln(1+\sqrt2)-\frac1{\sqrt2}-\frac1{6\sqrt2}-\frac1{20\sqrt2}.\)

So

\(I_3=\ln(1+\sqrt2)-\frac1{\sqrt2}\left(1+\frac16+\frac1{20}\right)=\ln(1+\sqrt2)-\frac1{\sqrt2}\cdot\frac{73}{60}.\)

Therefore

\(\int_0^{\pi/4}\frac{\sin^6 x}{\cos x}\,dx=\ln(1+\sqrt2)-\frac{73\sqrt2}{120}.\)