<p>The value of \ \int \cos^{\frac{1}{2}} x \cdot \sin^3 x \, dx\ is _______.</p>
Step-by-Step Solution
Key Concept: Rewrite sin³x as sin²x·sinx = (1-cos²x)·sinx, then use substitution u = cosx to convert the integral into a polynomial form in u.
<p><strong>Step 1:</strong> Rewrite sin³x as sin²x·sinx = (1 - cos²x)·sinx</p><p>∫cos^(1/2)x · sin³x dx = ∫cos^(1/2)x · (1 - cos²x)·sinx dx</p><p><strong>Step 2:</strong> Let u = cosx, then du = -sinx dx, so sinx dx = -du</p><p>= ∫u^(1/2)·(1 - u²)·(-du)</p><p>= -∫(u^(1/2) - u^(5/2)) du</p><p><strong>Step 3:</strong> Integrate term by term:</p><p>= -[u^(3/2)/(3/2) - u^(7/2)/(7/2)] + C</p><p>= -[2u^(3/2)/3 - 2u^(7/2)/7] + C</p><p>= -2u^(3/2)/3 + 2u^(7/2)/7 + C</p><p><strong>Step 4:</strong> Substitute back u = cosx:</p><p>= 2cos^(7/2)x/7 - 2cos^(3/2)x/3 + C</p><p>or equivalently: <strong>2/7·cos^(7/2)x - 2/3·cos^(3/2)x + C</strong></p><p>∴ Answer: <strong>2cos^(7/2)x/7 - 2cos^(3/2)x/3 + C</strong></p>
Correct Answer: 2