Question:

Compute the antiderivative of the following function:

Answer:

To find the antiderivative of $f(x) = \frac{\sin^3(x)}{\cos^2(x)}$, we can apply several techniques of integration. In this case, we will use a substitution to simplify the integral before proceeding with integration by parts.

First, we make the substitution $u = \cos(x)$, which implies $du = -\sin(x) \, dx$.

Next, we need to express the original function $f(x)$ in terms of $u$. Using the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$, we can rewrite $\sin^3(x)$ as $\sin^2(x) \cdot \sin(x) = (1-\cos^2(x))\sin(x) = (1-u^2)\sqrt{1-u^2}$ (since $\sin(x) = \sqrt{1-\cos^2(x)}$).

By substituting $(1-u^2)\sqrt{1-u^2}$ for $\sin^3(x)$ and $du = -\sin(x) \, dx$ in the original function, the integral becomes:

We can now simplify this expression further. Expanding the numerator and canceling terms with $u^2$, we have:

Splitting this into two separate integrals, we get:

We will evaluate each integral separately.

(i) Evaluating the integral $\int \frac{1}{u^2}\sqrt{1-u^2} \, du$:

We can rewrite this integral as:

To solve this integral, we can use the substitution $v = 1-u^2$. This implies $\frac{dv}{du} = -2u$ or $du = -\frac{dv}{2u}$.

Substituting, the integral becomes:

Integrating, we have:

Substituting $v = 1-u^2$, we get:

This integral represents a half of the area of a unit circle. It is a known result that $\int \sqrt{1-x^2} \, dx = \frac{1}{2}\sin^{-1}(x) + C_2$.

However, in our case, we have made the substitution $u = \cos(x)$. This implies $x = \cos^{-1}(u)$. Hence, the integral becomes:

Finally, combining the results of the two integrals, the antiderivative of $f(x) = \frac{\sin^3(x)}{\cos^2(x)}$ is given by:

where $C = C_1 + C_2$ represents the constant of integration.