Definite Integration
Grade 12

Question:

<p>Evaluate \(\displaystyle\int_0^{\pi/2} \sin^5 x\cos^4 x\,dx\)</p>
\(\dfrac{8}{630}\)
\(\dfrac{8}{315}\)
\(\dfrac{4}{315}\)
\(\dfrac{16}{315}\)

Step-by-Step Solution

Key Concept: Wallis formula: \int_0^(\pi/2) sinᵐx \cdot cosⁿx dx = [(m-1)\!\! \cdot (n-1)\!\!] / (m+n)\!\! \cdot factor
<div class='solution'> <p><strong>Using Wallis formula:</strong> \(\displaystyle\int_0^{\pi/2}\sin^m x\cos^n x\,dx = \frac{(m-1)\!\!(n-1)\!\!}{(m+n)\!\!}\) when \(m+n\) is odd, no extra factor.</p> <p>Here \(m=5\), \(n=4\): \(m+n=9\) (odd) \to no \(\pi/2\) factor.</p> <p>\((m-1)\!\! = 4\!\! = 4\cdot2 = 8\)</p> <p>\((n-1)\!\! = 3\!\! = 3\cdot1 = 3\)</p> <p>\((m+n)\!\! = 9\!\! = 9\cdot7\cdot5\cdot3\cdot1 = 945\)</p> <p>\[I = \frac{8\cdot3}{945} = \frac{24}{945} = \frac{8}{315}\]</p> <p><strong>Verification by substitution:</strong> Let \(t=\cos x\):</p> <p>\[I = \int_0^1(1-t^2)^2 t^4\,dt = \int_0^1(t^4-2t^6+t^8)\,dt = \frac{1}{5}-\frac{2}{7}+\frac{1}{9} = \frac{63-90+35}{315} = \frac{8}{315}\checkmark\]</p>
Correct Answer: B

Master Definite Integration with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free