Answer: (a) Let \(I_n=\int_0^{\pi/2}\sin^n\theta\,d\theta\). Then
\(I_n=\int_0^{\pi/2}\sin^{n+2}\theta\,\csc^2\theta\,d\theta\).
Integrating by parts with \(u=\sin^{n+2}\theta\) and \(dv=\csc^2\theta\,d\theta\), so that \(du=(n+2)\sin^{n+1}\theta\cos\theta\,d\theta\) and \(v=-\cot\theta\), gives
\(I_n=\left[-\sin^{n+2}\theta\cot\theta\right]_0^{\pi/2}+(n+2)\int_0^{\pi/2}\sin^{n+1}\theta\cos\theta\cot\theta\,d\theta\).
The boundary term is zero, and \(\cos\theta\cot\theta=\cos^2\theta/\sin\theta=1-\sin^2\theta\over\sin\theta\), so
\(I_n=(n+2)\int_0^{\pi/2}\sin^n\theta\,(1-\sin^2\theta)\,d\theta=(n+2)I_n-(n+2)I_{n+2}\).
Hence \(0=(n+1)I_n-(n+2)I_{n+2}\), so
\(I_{n+2}=\frac{n+1}{n+2}I_n\).
(b) The region has area \(A=\int_0^{\pi/(2m)}\sin^4(mx)\,dx\), and the centroid formula gives
\(\bar y=\dfrac{\frac12\int_0^{\pi/(2m)}\sin^8(mx)\,dx}{\int_0^{\pi/(2m)}\sin^4(mx)\,dx}.\)
With \(u=mx\),
\(\bar y=\dfrac{\frac12\int_0^{\pi/2}\sin^8u\,du}{\int_0^{\pi/2}\sin^4u\,du}=\frac12\,\frac{I_8}{I_4}.\)
Using the reduction formula:
\(I_0=\frac\pi2\), \(I_2=\frac12 I_0=\frac\pi4\), \(I_4=\frac34 I_2=\frac{3\pi}{16}\),
\(I_6=\frac56 I_4\), so \(I_8=\frac78 I_6=\frac78\cdot\frac56\cdot\frac34\cdot\frac\pi2=\frac{105\pi}{768}\).
Therefore
\(\bar y=\frac12\cdot\frac{I_8}{I_4}=\frac12\cdot\frac{105\pi/768}{3\pi/16}=\frac{35}{96}.\)
So the \(y\)-coordinate of the centroid is \(\frac{35}{96}\).
(a) Define \(I_n=\int_0^{\pi/2}\sin^n\theta\,d\theta\). Write
\(I_n=\int_0^{\pi/2}\sin^{n+2}\theta\,\csc^2\theta\,d\theta\).
Now integrate by parts, taking \(u=\sin^{n+2}\theta\) and \(dv=\csc^2\theta\,d\theta\). Then \(du=(n+2)\sin^{n+1}\theta\cos\theta\,d\theta\) and \(v=-\cot\theta\). So
\(I_n=\left[-\sin^{n+2}\theta\cot\theta\right]_0^{\pi/2}+(n+2)\int_0^{\pi/2}\sin^{n+1}\theta\cos\theta\cot\theta\,d\theta\).
The boundary term is zero, since \(\cot(\pi/2)=0\) and \(\sin 0=0\). Also \(\cos\theta\cot\theta=\cos^2\theta/\sin\theta\), so
\(\sin^{n+1}\theta\cos\theta\cot\theta=\sin^n\theta\cos^2\theta=\sin^n\theta\bigl(1-\sin^2\theta\bigr)\).
Hence
\(I_n=(n+2)\int_0^{\pi/2}\sin^n\theta\,d\theta-(n+2)\int_0^{\pi/2}\sin^{n+2}\theta\,d\theta\)
\(=(n+2)I_n-(n+2)I_{n+2}.\)
Rearranging gives
\(0=(n+1)I_n-(n+2)I_{n+2}\),
so
\(I_{n+2}=\frac{n+1}{n+2}I_n.\)
(b) For the region under \(y=\sin^4(mx)\) from \(x=0\) to \(x=\pi/(2m)\), the area is
\(A=\int_0^{\pi/(2m)}\sin^4(mx)\,dx.\)
The centroid formula for the \(y\)-coordinate is
\(\bar y=\frac{1}{A}\int_0^{\pi/(2m)}\frac{1}{2}\bigl(\sin^4(mx)\bigr)^2\,dx =\frac{\frac12\int_0^{\pi/(2m)}\sin^8(mx)\,dx}{\int_0^{\pi/(2m)}\sin^4(mx)\,dx}.\)
Let \(u=mx\). Then \(dx=du/m\), and the factor \(1/m\) cancels:
\(\bar y=\frac{\frac12\int_0^{\pi/2}\sin^8u\,du}{\int_0^{\pi/2}\sin^4u\,du}=\frac12\,\frac{I_8}{I_4}.\)
Now use the reduction formula repeatedly:
\(I_0=\int_0^{\pi/2}1\,d\theta=\frac\pi2\).
\(I_2=\frac{1}{2}I_0=\frac\pi4\).
\(I_4=\frac{3}{4}I_2=\frac{3\pi}{16}.\)
Then
\(I_6=\frac{5}{6}I_4\), and
\(I_8=\frac{7}{8}I_6=\frac78\cdot\frac56\cdot\frac34\cdot\frac\pi2=\frac{105\pi}{768}.\)
Therefore
\(\bar y=\frac12\cdot\frac{I_8}{I_4}=\frac12\cdot\frac{105\pi/768}{3\pi/16} =\frac12\cdot\frac{35}{24}=\frac{35}{96}.\)
So the \(y\)-coordinate of the centroid is \(\frac{35}{96}\).
