Answer: For the recurrence, write
\(I_{n+1}=\int_{0}^{\pi/4}\frac{\sin^{2n+2}x}{\cos x}\,dx\).
Now use \(\sin^2 x=1-\cos^2 x\):
\(I_{n+1}=\int_{0}^{\pi/4}\frac{(1-\cos^2 x)\sin^{2n}x}{\cos x}\,dx\)
\(=\int_{0}^{\pi/4}\frac{\sin^{2n}x}{\cos x}\,dx-\int_{0}^{\pi/4}\cos x\sin^{2n}x\,dx\)
\(=I_n-\int_{0}^{\pi/4}\cos x\sin^{2n}x\,dx\).
Since \(\dfrac{d}{dx}(\sin^{2n+1}x)=(2n+1)\sin^{2n}x\cos x\),
\(\int_{0}^{\pi/4}\cos x\sin^{2n}x\,dx=\left[\frac{\sin^{2n+1}x}{2n+1}\right]_{0}^{\pi/4}=\frac{(\sin\frac{\pi}{4})^{2n+1}}{2n+1}=\frac{2^{-(n+\frac12)}}{2n+1}.\)
Hence
\(I_n-I_{n+1}=\frac{2^{-(n+\frac12)}}{2n+1}.\)
To find \(\displaystyle \int_{0}^{\pi/4}\frac{\sin^6 x}{\cos x}\,dx\), note that this is \(I_3\). First,
\(I_0=\int_0^{\pi/4}\sec x\,dx=[\ln|\sec x+\tan x|]_0^{\pi/4}=\ln(1+\sqrt2).\)
Then use the recurrence:
\(I_0-I_1=\frac{2^{-1/2}}{1}=\frac1{\sqrt2}\), so \(I_1=\ln(1+\sqrt2)-\frac1{\sqrt2}\).
\(I_1-I_2=\frac{2^{-3/2}}{3}=\frac1{6\sqrt2}\), so \(I_2=\ln(1+\sqrt2)-\frac1{\sqrt2}-\frac1{6\sqrt2}\).
\(I_2-I_3=\frac{2^{-5/2}}{5}=\frac1{20\sqrt2}\), so
\(I_3=\ln(1+\sqrt2)-\frac1{\sqrt2}-\frac1{6\sqrt2}-\frac1{20\sqrt2}.\)
Now
\(\frac1{\sqrt2}+\frac1{6\sqrt2}+\frac1{20\sqrt2}=\frac{1}{\sqrt2}\left(1+\frac16+\frac1{20}\right)=\frac{73}{60\sqrt2}=\frac{73\sqrt2}{120}.\)
Therefore \(\displaystyle \int_{0}^{\pi/4}\frac{\sin^6 x}{\cos x}\,dx=\ln(1+\sqrt2)-\frac{73\sqrt2}{120}.\)
(a) Let \(I_n=\int_{0}^{\pi/4}\frac{\sin^{2n}x}{\cos x}\,dx\). Then
\(I_{n+1}=\int_{0}^{\pi/4}\frac{\sin^{2n+2}x}{\cos x}\,dx\)
\(=\int_{0}^{\pi/4}\frac{(1-\cos^2x)\sin^{2n}x}{\cos x}\,dx\)
\(=\int_{0}^{\pi/4}\frac{\sin^{2n}x}{\cos x}\,dx-\int_{0}^{\pi/4}\cos x\sin^{2n}x\,dx\)
\(=I_n-\int_{0}^{\pi/4}\cos x\sin^{2n}x\,dx\).
Now integrate \(\cos x\sin^{2n}x\) by recognising it as a derivative:
\(\dfrac{d}{dx}(\sin^{2n+1}x)=(2n+1)\sin^{2n}x\cos x\),
so
\(\int_{0}^{\pi/4}\cos x\sin^{2n}x\,dx=\left[\frac{\sin^{2n+1}x}{2n+1}\right]_{0}^{\pi/4}=\frac{(\sin\frac{\pi}{4})^{2n+1}}{2n+1}\)
\(=\frac{(\frac{1}{\sqrt2})^{2n+1}}{2n+1}=\frac{2^{-(n+\frac12)}}{2n+1}.\)
Hence
\(I_n-I_{n+1}=\frac{2^{-(n+\frac12)}}{2n+1}.\)
(b) We want \(\int_{0}^{\pi/4}\frac{\sin^6x}{\cos x}\,dx=I_3\). First,
\(I_0=\int_{0}^{\pi/4}\sec x\,dx=[\ln|\sec x+\tan x|]_{0}^{\pi/4}=\ln(1+\sqrt2).\)
Using the recurrence with \(n=0,1,2\):
\(I_0-I_1=\frac{2^{-1/2}}{1}=\frac1{\sqrt2}\), so \(I_1=\ln(1+\sqrt2)-\frac1{\sqrt2}\).
\(I_1-I_2=\frac{2^{-3/2}}{3}=\frac1{6\sqrt2}\), so \(I_2=\ln(1+\sqrt2)-\frac1{\sqrt2}-\frac1{6\sqrt2}\).
\(I_2-I_3=\frac{2^{-5/2}}{5}=\frac1{20\sqrt2}\), so
\(I_3=\ln(1+\sqrt2)-\frac1{\sqrt2}-\frac1{6\sqrt2}-\frac1{20\sqrt2}.\)
Combine the fractional terms:
\(\frac1{\sqrt2}+\frac1{6\sqrt2}+\frac1{20\sqrt2}=\frac{1}{\sqrt2}\left(1+\frac16+\frac1{20}\right)=\frac{73}{60\sqrt2}=\frac{73\sqrt2}{120}.\)
Therefore
\(\int_{0}^{\pi/4}\frac{\sin^6x}{\cos x}\,dx=\ln(1+\sqrt2)-\frac{73\sqrt2}{120}.\)
