Answer: \(y=-\dfrac{1}{16}\cos\!\left(4x-\dfrac{\pi}{4}\right)+\dfrac12x+\dfrac{5\pi}{32}\).
We start with the main method. Integrate using the reverse chain rule, taking care with the multiplier inside the trigonometric function.
Integrate the second derivative once:
\(\frac{\mathrm dy}{\mathrm dx}=\frac14\sin\left(4x-\frac{\pi}{4}\right)+C.\)
At \(x=\frac{3\pi}{16}\),
\(4x-\frac{\pi}{4}=\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}.\)
Since \(\frac{\mathrm dy}{\mathrm dx}=\frac34\) there,
\(\frac34=\frac14\sin\frac{\pi}{2}+C.\)
So \(C=\frac12\). Hence
\(\frac{\mathrm dy}{\mathrm dx}=\frac14\sin\left(4x-\frac{\pi}{4}\right)+\frac12.\)
Integrate again:
\(y=-\frac1{16}\cos\left(4x-\frac{\pi}{4}\right)+\frac12x+A.\)
The point \(\left(\frac{3\pi}{16},\frac{\pi}{4}\right)\) lies on the curve, so
\(\frac{\pi}{4}=-\frac1{16}\cos\frac{\pi}{2}+\frac12\cdot\frac{3\pi}{16}+A.\)
Since \(\cos\frac{\pi}{2}=0\),
\(\frac{\pi}{4}=\frac{3\pi}{32}+A.\)
Therefore \(A=\frac{5\pi}{32}\), and the equation of the curve is
\(y=-\frac1{16}\cos\left(4x-\frac{\pi}{4}\right)+\frac12x+\frac{5\pi}{32}.\)
This completes the solution and gives the required result.
