Question

If \(\int \sin ^{4}x\cdot e^{\log (\cos x)}dx=\frac{1}{k}\sin ^{k}x+C\), for \(x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\), then ...

• A. \(k=4\)
• B. \(k=0\)
• C. \(k=1\)
• D. \(k=5\)

Hint:

Use e^(log cos x) = cos x, then substitute t = sin x.

Answer 1

The Correct Option is D

Step 1: Differentiate the right side
$\dfrac{d}{dx}\left(\dfrac{1}{k}\sin ^{k}x\right)=\sin ^{k-1}x\cos x$.

Step 2: Compare with the integrand
The integrand is $\sin ^{4}x\cos x$, so $k-1=4$.

Step 3: Solve
Hence $k=5$.

Final Answer:
Matching the power of sine gives $k=5$. \[ \boxed{\text{(D) }k=5} \]