Solution (a)

Step 1 (Simplification): Use the logarithmic property $e^{\log f(x)} = f(x)$.

The integral becomes:

$$I = \int \cos^3 x \cdot \sin x \, dx$$

Step 2 (Substitution): Let $u = \cos x$. Then $du = -\sin x \, dx$, which implies $\sin x \, dx = -du$.

Substituting into the integral:

$$I = \int u^3 (-du) = -\int u^3 \, du$$

Step 3 (Final Answer): Integrating with respect to $u$ and substituting back:

$$I = -\frac{u^4}{4} + C = -\frac{\cos^4 x}{4} + C$$

Solution (b)

Step 1 (Completing the Square): Rewrite the denominator $5 + 4x - x^2$ by completing the square.

$$5 + 4x - x^2 = 5 - (x^2 - 4x) = 5 - (x^2 - 4x + 4 - 4) = 9 - (x - 2)^2$$

The integral becomes:

$$I = \int \frac{1}{3^2 - (x - 2)^2} \, dx$$

Step 2 (Formula Application): Use the standard integral formula $\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$.

Here, $a = 3$ and $x$ is replaced by $(x-2)$:

$$I = \frac{1}{2(3)} \log \left| \frac{3 + (x - 2)}{3 - (x - 2)} \right| + C$$

Step 3 (Final Answer): Simplify the expression:

$$I = \frac{1}{6} \log \left| \frac{x + 1}{5 - x} \right| + C$$