Definite Integration
Grade None
Question:
<p>The integral <span class="math-tex">\(\int_{0}^{\pi} \frac{8 x d x}{4 \cos ^{2} x+\sin ^{2} x}\)</span> is equal to</p>
<p style="display:inline"><span class="math-tex">\(\pi^{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \pi^{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3 \pi^{2}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(4 \pi^{2}\)</span></p>
Step-by-Step Solution
Key Concept: Use the 'King's Rule' property to eliminate the variable x in the numerator and then apply the periodic property to change the upper limit to π/2 before substituting t = tan x.
<p>Let <span class="math-tex">$I=\int_{0}^{\pi} \frac{8 x d x}{4 \cos ^{2} x+\sin ^{2} x}$</span> ...(i)<br />
<span class="math-tex">$I=\int_{0}^{\pi} \frac{8(\pi-x) d x}{4 \cos ^{2}(\pi-x)+\sin ^{2}(\pi-x)}$</span> <span class="math-tex">$\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$</span><br />
<span class="math-tex">$\therefore I=\int_{0}^{\pi} \frac{8(\pi-x)}{4 \cos ^{2} x+\sin ^{2} x} d x$</span> ...(ii)<br />
Adding eqn (i) and (ii), we get<br />
<span class="math-tex">$2 I=8 \pi \int_{0}^{\pi} \frac{d x}{4 \cos ^{2} x+\sin ^{2} x}$</span><br />
<span class="math-tex">$2 I=8 \pi \times 2 \int_{0}^{\pi / 2} \frac{\sec ^{2} x}{4+\tan ^{2} x} d x$</span><br />
<span class="math-tex">$I=8 \pi \int_{0}^{\infty} \frac{d t}{4+t^{2}}=8 \pi \times \frac{1}{2}\left[\tan ^{-1} \frac{t}{2}\right]_{0}^{\infty}$</span><br />
<span class="math-tex">$=4 \pi \times \frac{\pi}{2}=2 \pi^{2}$</span></p>
Correct Answer: B