Question

$\int \cos^3 x \cdot e^{\log(\sin x)} , dx = $

• A. $-\frac{e^{\sin x}}{4} + c$
• B. $-\frac{\cos^4 x}{4} + c$
• C. $-\frac{\sin^4 x}{4} + c$
• D. $\frac{e^{\sin x}}{4} + c$

Hint:

Whenever an integrand contains an expression of the form $f'(x)[f(x)]^n$, its integrated solution is always $\frac{[f(x)]^{n+1}}{n+1} + c$. Recognizing that $-\sin x$ is the direct derivative of $\cos x$ allows you to evaluate this integral mentally in one step!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem requires evaluating the indefinite integral of the compound trigonometric exponential function $\cos^3 x \cdot e^{\log(\sin x)}$.

Step 2: Key Formula or Approach:
First, simplify the integrand using the fundamental logarithmic identity relating exponential and logarithmic inverses: $$e^{\log(f(x))} = f(x)$$ This simplification transforms the integral into a standard format where we can easily apply the method of integration by substitution.

Step 3: Detailed Explanation:
Apply the identity $e^{\log(\sin x)} = \sin x$ to simplify our integral expression: $$I = \int \cos^3 x \cdot \sin x \, dx$$ Now use the substitution method. Let $t = \cos x$. Differentiate both sides to find $dt$: $$\frac{dt}{dx} = -\sin x \implies \sin x \, dx = -dt$$ Substitute these new variables into our simplified integral: $$I = \int t^3 \cdot (-dt) = -\int t^3 \, dt$$ Evaluate the integral using the power rule $\int t^n \, dt = \frac{t^{n+1}}{n+1}$: $$I = -\frac{t^4}{4} + c$$ Substitute back the original variable $t = \cos x$ to arrive at the final answer: $$I = -\frac{\cos^4 x}{4} + c$$ This matches the expression in option (B).

Step 4: Final Answer:
The value of the indefinite integral is $-\frac{\cos^4 x}{4} + c$, which corresponds to option (B).