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: Understanding the Concept
For \(x\in \left(-\dfrac{\pi }{2},\dfrac{\pi }{2}\right)\), \(\cos x>0\), so \(e^{\log (\cos x)}=\cos x\).

Step 2: Rewrite
The integral becomes \(\int \sin ^{4}x\cos x\,dx\).

Step 3: Substitute
Let \(t=\sin x\), so \(dt=\cos x\,dx\). Then \(\int t^{4}dt=\dfrac{t^{5}}{5}+C=\dfrac{1}{5}\sin ^{5}x+C\).

Step 4: Compare
This matches \(\dfrac{1}{k}\sin ^{k}x+C\) with \(k=5\).

Final Answer:
The value of \(k\) is \(5\). \[ \boxed{\text{(D) }k=5} \]