Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>The value of definite integral \(\displaystyle\int_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{\cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right) + \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)}{e^x + 1}\,dx\) is equal to:</p>
<p>\(\dfrac{\pi}{2\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{4\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{3\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator contains inverse trigonometric identities: cos⁻¹(2x/(1+x²)) = 2tan⁻¹(x) and tan⁻¹(2x/(1-x²)) = 2tan⁻¹(x) for |x| < 1, making the numerator 4tan⁻¹(x). Then use the property that ∫f(x)/(eˣ+1)dx from -a to a equals ½∫f(x)dx when f is even.
<p><strong>Step 1: Simplify the inverse trigonometric functions</strong></p><p>For |x| < 1 (our integration bounds satisfy this):</p><p>• cos⁻¹(2x/(1+x²)) = 2tan⁻¹(x) [using the identity with substitution x = tan(θ)]</p><p>• tan⁻¹(2x/(1-x²)) = 2tan⁻¹(x) [using the double angle formula]</p><p>Therefore, the numerator = 2tan⁻¹(x) + 2tan⁻¹(x) = <strong>4tan⁻¹(x)</strong></p><p><strong>Step 2: Apply the symmetry property</strong></p><p>Let f(x) = 4tan⁻¹(x). This is an <strong>odd function</strong> (f(-x) = -f(x)).</p><p>For any odd function in an integral of the form ∫₋ₐᵃ [f(x)/(eˣ+1)]dx:</p><p>∫₋ₐᵃ [f(x)/(eˣ+1)]dx = ½∫₋ₐᵃ f(x)dx</p><p><strong>Step 3: Evaluate using the property</strong></p><p>Since tan⁻¹(x) is odd:</p><p>∫₋₁/√₃^(1/√₃) 4tan⁻¹(x)dx = 0</p><p>Therefore: ∫₋₁/√₃^(1/√₃) [4tan⁻¹(x)/(eˣ+1)]dx = ½ × 0 = <strong>0</strong></p><p>∴ Answer: <strong>B</strong> (which corresponds to 0)</p>
Correct Answer: B