Question

1. \\( \frac { d y } { d u } = 4 ( \sin u ) ^ { 3 } ( \cos u ) \\)

Explanation:

Step1: Integrate both sides

$$y=\int4(\sin u)^3(\cos u)du$$
Let \(t = \sin u\), then \(dt=\cos udu\).

Step2: Substitute \(t\)

The integral becomes \(y = 4\int t^3dt\).
Using the power - rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 3\), we have \(4\times\frac{t^{4}}{4}+C\).

Step3: Substitute back \(t=\sin u\)

\(y=\sin^{4}u + C\)

Answer:

\(y=\sin^{4}u + C\)