Question

Multiple Choice If \(\int x^{2} \cos x d x=h(x)-\int 2 x \sin x d x,\) then \(h(x)=\) (A) \(2 \sin x+2 x \cos x+C\) (B) \(x^{2} \sin x+C\) (C) \(2 x \cos x-x^{2} \sin x+C\) (D) \(4 \cos x-2 x \sin x+C\) (E) \(\left(2-x^{2}\right) \cos x-4 \sin x+C\)

Short answer

The solution is (B) \(x^2 sin(x) + C\).

Step-by-step solution

Identify the parts

First, let's identify which parts of the given integral follow the formula for integration by parts. We have \(u = x^2\), \(dv = cos(x)dx\). To complete the formula we need to find \(du\) and \(v\). The derivative of \(x^2\) is \(2x\), so \(du = 2x dx\). And the integral of \(cos(x)\) is \(sin(x)\), so \(v = sin(x)\).

Apply the integration by parts formula

The integration by parts formula is: \(\int u dv = uv - \int v du\). Substituting the values we identified, we get \(\int x^2 cos(x) dx = x^2 sin(x) - \int 2x sin(x) dx\). The left-hand side of this equation is equal to \(h(x) - \int 2x sin(x) dx\). If we set these two expressions equal, we can solve for \(h(x)\).

Solve for h(x)

Setting the two expressions equal, we have \(h(x) - \int 2x sin(x) dx = x^2 sin(x) - \int 2x sin(x) dx\). Simplifying, \(\int 2x sin(x) dx\) cancels out on both sides and we're left with \(h(x) = x^2 sin(x) + C\).