Question

In Exercises \(21-26,\) construct a function of the form \(y=\int^{x} f(t) d t+C\) that satisfies the given conditions. $$\frac{d y}{d x}=\sin ^{3} x,\( and \)y=0\( when \)x=5$$

Short answer

The function is \(y = \int^{x} \sin^{3} t dt + C\) where \(C = [0 - \int^{5} \sin^{3} t dt]\).

Step-by-step solution

Derive the function

The function can be derived based on \(f(t) = \sin^{3} x\). It is given that \(y = \int^{x} f(t)dt + C\), replacing the \(f(t)\) with \(\sin^{3} x\), the function becomes \(y = \int^{x} \sin^{3} t dt + C\).

Set up the equation to solve for C

We know that \(y\) equals \(0\) when \(x = 5\). Replacing these values in the function, \(0 = \int^{5} \sin^{3} t dt + C\). This can now be solved for \(C\).

Solve for C

To solve for the constant \(C\) consider \(\int^{5} \sin^3 t dt\). The integrand \(\sin^3 t\) is an odd-numbered power of sine, and its integral can be computed by saving a factor of \(\sin t\), converting the remaining \(\sin^2 t\) to \(\cos^2 t\) via the Pythagorean identity, and then applying a power-reducing formula. This process yields \(-1/3 cos t * (1 - 1/2 cos^2 t)^2 |_0^5\). Evaluate this integral from \(0\) to \(5\) and we'll have the value of \(-1/3 cos t * (1 - 1/2 cos^2 t)^2 |_0^5\). Now, subtract this value from both sides of the equation obtained in step 2 to solve for \(C\).