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We are given the integral $\int \cos^3 x \cdot e^{\log(\sin x)} dx$. Since $e^{\log a} = a$, we can simplify the integrand as follows: $$e^{\log(\sin x)} = \sin x$$ Thus, the integral becomes: $$\int \cos^3 x \sin x dx$$
Let $u = \cos x$. Then, $\frac{du}{dx} = -\sin x$, which implies $du = -\sin x dx$. So, $\sin x dx = -du$. Substituting these into the integral, we get: $$\int \cos^3 x \sin x dx = \int u^3 (-du) = -\int u^3 du$$
Now, we can integrate $u^3$ with respect to $u$: $$-\int u^3 du = -\frac{u^4}{4} + C$$ where $C$ is the constant of integration.
Substitute $u = \cos x$ back into the expression: $$-\frac{u^4}{4} + C = -\frac{\cos^4 x}{4} + C$$ Thus, the integral is: $$\int \cos^3 x \sin x dx = -\frac{\cos^4 x}{4} + C$$
Final Answer: $-\frac{\cos^4 x}{4} + C$
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