First slide

Evaluation of definite integrals

Question

The integral $\int_{\sqrt{\log 5}}^{\sqrt{\log 7}} \frac{x \cos x^{2}}{\cos (\log 35 - x^{2}) + \cos x^{2}} d x$ is
equal to

Moderate

A

$\frac{1}{4} \log \frac{5}{7}$
B

$\frac{1}{2} \log \frac{5}{7}$
C

$\frac{1}{4} \log \frac{7}{5}$
D

$\frac{1}{2} \log \frac{7}{5}$

Solution

$\begin{array}{l} \int_{\sqrt{\log 5}}^{\sqrt{\log 7}} \frac{x \cos x^{2}}{\cos (\log 35 - x^{2}) + \cos x^{2}} d x \\ = \frac{1}{2} \int_{\log 5}^{\log 7} \frac{\cos t}{\cos (\log 35 - t) + \cos t} d t (x^{2} = t) \\ = \frac{1}{2} I^{'} \end{array}$
$I^{'} = \int_{\log 5}^{\log 7} \frac{\cos (\log 7 + \log 5 - t)}{\cos t + \cos (\log 7 + \log 5 - t)} d t$
$= \int_{\log 5}^{\log 7} \frac{\cos (\log 35 - t)}{\cos t + \cos (\log 35 - t)} d t$
$\begin{array}{l} 2 I^{'} = I^{'} + I^{'} = \int_{\log 5}^{\log 7} \frac{\cos t + \cos (\log 35 - t)}{\cos t + \cos (\log 35 - t)} d t \\ = \log \frac{7}{5} \\ \Rightarrow I = \frac{1}{4} \log \frac{7}{5} \end{array}$

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