First slide

Evaluation of definite integrals

Question

The value of $I = \int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^{2}}{\sin x^{2} + \sin (\log 6 - x^{2})} d x$ is

Easy

A

$\frac{1}{4} \log \frac{3}{2}$
B

$\frac{1}{2} \log \frac{3}{2}$
C

$\log \frac{3}{2}$
D

$\frac{1}{6} \log \frac{3}{2}$

Solution

Put $x^{2} = t$ , so that
$\begin{array}{l} I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t}{\sin t + \sin (\log 6 - t)} d t \\ = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2 + \log 3 - t)}{\sin (\log 2 + \log 3 - t) + \sin t} d t \\ = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6 - t)}{\sin t + \sin (\log 6 - t)} d t \\ 2 I = I + I \end{array}$
$\begin{array}{l} = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t + \sin (\log 6 - t)}{\sin t + \sin (\log 6 - t)} d t \\ = \frac{1}{2} [\log 3 - \log 2] \end{array}$
$\Rightarrow I = \frac{1}{4} \log \frac{3}{2}$ .

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