Question:
The value of \(\displaystyle\int_0^{\pi/2} \sin^2(x)\cos^4(x)\,dx\) is ____.
The value of \(\displaystyle\int_0^{\pi/2} \sin^2(x)\cos^4(x)\,dx\) is ____.
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Write \(\sin^2x\cos^4x=\cos^4x-\cos^6x\) and use Wallis' formula.
Updated On: Jul 3, 2026
- \(\dfrac{\pi}{24}\)
- \(\dfrac{\pi}{32}\)
- \(\dfrac{2\pi}{15}\)
- \(\dfrac{3\pi}{32}\)
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The Correct Option is B
Solution and Explanation
Step 1: Rewrite the integrand using \(\sin^2x = 1-\cos^2x\).
\[
\sin^2x\cos^4x = \cos^4x - \cos^6x
\]
So the integral splits as
\[
I = \int_0^{\pi/2}\cos^4x\,dx - \int_0^{\pi/2}\cos^6x\,dx
\]
Step 2: Evaluate each piece with Wallis' formula \(\displaystyle\int_0^{\pi/2}\cos^{2n}x\,dx = \frac{(2n-1)!!}{(2n)!!}\cdot\frac{\pi}{2}\).
For \(n=2\):
\[
\int_0^{\pi/2}\cos^4x\,dx = \frac{3\cdot1}{4\cdot2}\cdot\frac{\pi}{2} = \frac{3\pi}{16}
\]
For \(n=3\):
\[
\int_0^{\pi/2}\cos^6x\,dx = \frac{5\cdot3\cdot1}{6\cdot4\cdot2}\cdot\frac{\pi}{2} = \frac{15}{48}\cdot\frac{\pi}{2} = \frac{5\pi}{32}
\]
Step 3: Subtract the two results.
\[
I = \frac{3\pi}{16} - \frac{5\pi}{32} = \frac{6\pi}{32} - \frac{5\pi}{32} = \frac{\pi}{32}
\]
\[
\boxed{I = \dfrac{\pi}{32}}
\]
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