Indefinite Integration
Trigonometric Integrals
Grade 12
Question:
<p>The integral \(\int \sqrt{1 + 2\cot x(\csc x + \cot x)}\, dx\) is equal to (where \(C\) is a constant of integration):</p>
<p>\(2\log\left|\sin\dfrac{x}{2}\right| + C\)</p>
<p>\(2\log\left|\cos\dfrac{x}{2}\right| + C\)</p>
<p>\(4\log\left|\sin\dfrac{x}{2}\right| + C\)</p>
<p>\(4\log\left|\cos\dfrac{x}{2}\right| + C\)</p>
Step-by-Step Solution
Key Concept: Recognize that the expression under the square root can be rewritten as a perfect square by simplifying 1 + 2cot x(csc x + cot x) = (csc x + cot x)², then use the substitution property √(u²) = |u| to integrate.
<p><strong>Step 1:</strong> Simplify the expression under the square root.</p><p>1 + 2cot x(csc x + cot x) = 1 + 2cot x·csc x + 2cot²x</p><p><strong>Step 2:</strong> Recognize this as a perfect square.</p><p>Note that (csc x + cot x)² = csc²x + 2csc x·cot x + cot²x = (1 + cot²x) + 2csc x·cot x + cot²x = 1 + 2csc x·cot x + 2cot²x</p><p>Therefore: √(1 + 2cot x·csc x + 2cot²x) = |csc x + cot x|</p><p><strong>Step 3:</strong> Integrate using the standard formula.</p><p>∫|csc x + cot x| dx = -ln|csc x + cot x| + C</p><p>Or equivalently: ∫(csc x + cot x) dx = -ln|csc x + cot x| + C = ln|csc x - cot x| + C</p><p>∴ Answer: A</p>
Correct Answer: A