Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>The integral <equation>\int \frac{(2x - 1) \cos\sqrt{(2x-1)^2 + 5}}{4x^2 - 4x + 6} dx</equation> is equal to (where C is a constant of integration) (JEE Main 2021)</p>
<p>(a) <equation>\sin\sqrt{(2x-1)^2 + 5} + C</equation></p>
<p>(b) <equation>\frac{1}{2}\sin\sqrt{(2x-1)^2 + 5} + C</equation></p>
<p>(c) <equation>\cos\sqrt{(2x-1)^2 + 5} + C</equation></p>
<p>(d) <equation>\frac{1}{2}\cos\sqrt{(2x-1)^2 + 5} + C</equation></p>
Step-by-Step Solution
Key Concept: Recognize that the derivative of the expression under the radical matches the numerator structure, allowing substitution.
<p><strong>Step 1:</strong> Notice that <equation>\frac{d}{dx}[(2x-1)^2 + 5] = 2(2x-1) \cdot 2 = 4(2x-1)</equation>.</p><p><strong>Step 2:</strong> Rewrite the denominator: <equation>4x^2 - 4x + 6 = (2x-1)^2 + 5</equation>.</p><p><strong>Step 3:</strong> Use substitution <equation>u = \sqrt{(2x-1)^2 + 5}</equation>, so <equation>du = \frac{2(2x-1)}{\sqrt{(2x-1)^2 + 5}} dx</equation>.</p><p><strong>Step 4:</strong> The integral becomes <equation>\int \cos u \cdot \frac{1}{2} du = \frac{1}{2}\sin u + C = \frac{1}{2}\sin\sqrt{(2x-1)^2 + 5} + C</equation>.</p><p>∴ Answer is (b).</p>
Correct Answer: B