Definite Integration
Grade None
Question:
<p>If the value of the integral <span class="math-tex">\(\int_{-1}^{1} \frac{\cos \alpha x}{1+3^{x}} d x\)</span> is <span class="math-tex">\(\frac{2}{\pi}\)</span>. Then, a value of <span class="math-tex">\(\alpha\)</span> is</p>
<p style="display:inline"><span class="math-tex">\(\frac{\pi}{6}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\pi}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\pi}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\pi}{2}\)</span></p>
Step-by-Step Solution
<p>Let <span class="math-tex">${I}=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^{x}} d x$</span> ...(i)<br />
<span class="math-tex">${I}=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^{-x}} d x$</span> ...(i)<br />
(Using <span class="math-tex">$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$</span>)<br />
Adding eqn. (i) and (ii)<br />
<span class="math-tex">$2 {I}=\int_{-1}^{+1} \cos (\alpha x) d x=$</span> <span class="math-tex">$2 \int_{0}^{1} \cos (\alpha x) d x$</span><br />
<span class="math-tex">$I=\frac{\sin \alpha}{\alpha}=\frac{2}{\pi}($</span>given)<br />
<span class="math-tex">$\therefore \alpha=\frac{\pi}{2}$</span></p>
Correct Answer: D