Indefinite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>If <span class="math-inline">\(I = \int_{-1}^{1} \frac{\cos^{-1}\left(\frac{x^4}{4}\right)}{1+x^2} dx = k\int_{0}^{1} \frac{\cos^{-1}\left(\frac{x^4}{4}\right)}{1+x^2} dx\)</span>, then find <span class="math-inline">\(k\)</span>.</p>
Step-by-Step Solution
Key Concept: Identify that the integrand is an even function and apply the property that the integral of an even function over a symmetric interval equals twice the integral over the positive half.
<p><strong>Step 1:</strong> Let <span class="math-inline">$I = \int_{-1}^{1} \frac{\cos^{-1}\left(\frac{x^4}{4}\right)}{1+x^2} dx$</span></p><p><strong>Step 2:</strong> Observe that the integrand <span class="math-inline">$f(x) = \frac{\cos^{-1}\left(\frac{x^4}{4}\right)}{1+x^2}$</span> is an even function since <span class="math-inline">$(-x)^4 = x^4$</span> and <span class="math-inline">$1+(-x)^2 = 1+x^2$</span>.</p><p><strong>Step 3:</strong> For an even function, <span class="math-inline">$\int_{-a}^{a} f(x)dx = 2\int_{0}^{a} f(x)dx$</span></p><p><strong>Step 4:</strong> Therefore, <span class="math-inline">$I = \int_{-1}^{1} \frac{\cos^{-1}\left(\frac{x^4}{4}\right)}{1+x^2} dx = 2\int_{0}^{1} \frac{\cos^{-1}\left(\frac{x^4}{4}\right)}{1+x^2} dx$</span></p><p><strong>Step 5:</strong> Comparing with the given form, <span class="math-inline">$k = 2$</span>.</p>
Correct Answer: 2