First slide

Evaluation of definite integrals

Question

The integral $\int_{0}^{1 / a} \frac{\log (1 + a x)}{1 + a^{2} x^{2}} d x (a > 0)$ is equal to

Moderate

A

$a (\log 2) \frac{π}{8}$
B

$\frac{1}{a} (\log 2) \frac{π}{4}$
C

$\frac{1}{a} (\log 2) \frac{π}{8}$
D

$\frac{1}{a^{2}} (\log 2)$

Solution

Put $a x = t,$ the given integral reduces to
$\begin{array}{l} \frac{1}{a} \int_{0}^{1} \frac{\log (1 + t)}{1 + t^{2}} d t = \frac{1}{a} \int_{0}^{π / 4} \frac{\log (1 + \tan u)}{1 + \tan^{2} u} \sec^{2} u d u (t - \tan u) \\ = \frac{1}{a} \int_{0}^{π / 4} \log (1 + \tan u) d u \\ = \frac{1}{a} \int_{0}^{π / 4} \log (1 + \tan (\frac{π}{4} - u)) d u \\ = \frac{1}{a} \int_{0}^{π / 4} \log (1 + \frac{1 - \tan u}{1 + \tan u}) d u \\ = \frac{1}{a} \int_{0}^{π / 4} [\log 2 - \log (1 + \tan u)] d u \\ \Rightarrow \frac{2}{a} \int_{0}^{π / 4} \log (1 + \tan u) d u = \frac{1}{a} (\log 2) \frac{π}{4} \\ \Rightarrow \frac{1}{a} \int_{0}^{π / 4} \log (1 + \tan u) d u = \frac{1}{a} (\log 2) \frac{π}{8} \end{array}$

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