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: Simplify the exponential.
Since $e^{\log(\sin x)} = \sin x$, the integral becomes $\displaystyle\int \cos^3 x \cdot \sin x\, dx$.
Step 2: Choose a substitution.
The factor $\cos^3 x$ pairs with the cosine base, so let $t = \cos x$.
Step 3: Differentiate.
Then $dt = -\sin x\, dx$, so $\sin x\, dx = -dt$.
Step 4: Rewrite the integral.
$\displaystyle\int t^3(-dt) = -\int t^3\, dt$.
Step 5: Apply the power rule.
$-\dfrac{t^4}{4} + c$.
Step 6: Return to $x$.
Replacing $t=\cos x$ gives $-\dfrac{\cos^4 x}{4} + c$. A quick differentiation check returns the original integrand, confirming the result.
\[ \boxed{-\dfrac{\cos^4 x}{4} + c} \]