To solve the integral

$\int_{-\frac{\pi}{2}}^0 2 \sin^2(x) \cos(x) \, dx,$

we can use a substitution method to simplify it.

Step 1: Choose a Substitution

Observe that the integrand contains $\sin^2(x)$ and $\cos(x)$. Let's set $u = \sin(x).$ Then, $\frac{du}{dx} = \cos(x) \quad \Rightarrow \quad du = \cos(x) \, dx.$

Step 2: Rewrite the Limits of Integration

When $x = -\frac{\pi}{2}$: $u = \sin\left(-\frac{\pi}{2}\right) = -1.$ When $x = 0$: $u = \sin(0) = 0.$ Thus, the integral limits change from $x \in \left[-\frac{\pi}{2}, 0\right]$ to $u \in [-1, 0]$.

Step 3: Substitute and Simplify the Integral

Substitute $\sin(x) = u$, $\cos(x) \, dx = du$, and $\sin^2(x) = u^2$: $\int_{-\frac{\pi}{2}}^0 2 \sin^2(x) \cos(x) \, dx = \int_{-1}^0 2 u^2 \, du.$

Step 4: Integrate

Now we integrate with respect to $u$: $\int_{-1}^0 2 u^2 \, du = 2 \int_{-1}^0 u^2 \, du.$ The antiderivative of $u^2$ is $\frac{u^3}{3}$, so we get: $2 \int_{-1}^0 u^2 \, du = 2 \cdot \frac{u^3}{3} \Big|_{-1}^0 = \frac{2}{3} \left(0^3 - (-1)^3\right).$ This simplifies to: $\frac{2}{3} (0 + 1) = \frac{2}{3}.$

Final Answer

$\int_{-\frac{\pi}{2}}^0 2 \sin^2(x) \cos(x) \, dx = \frac{2}{3}.$

Would you like further clarification on any part of this solution?

Here are some related questions you might find helpful:

1. How does choosing different substitutions affect integral solutions?
2. What other methods could be used to evaluate similar trigonometric integrals?
3. How do integral limits transform under substitution?
4. Why is the antiderivative of $u^2$ equal to $\frac{u^3}{3}$?
5. Can we solve this integral using trigonometric identities instead?

Tip: When choosing substitutions, look for parts of the integrand whose derivatives also appear in the expression, as this often simplifies the integral.