Question
Evaluate: $\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{3 \tan ^2 x+4 \tan x+1} d x$
✓
Answer
$\text { Let } I =\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{3 \tan ^2 x+4 \tan x+1} d x$
Put $\tan x=t$
$\therefore \sec ^2 x d x = dt$
When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$
$ \therefore I=\int_0^1 \frac{ dt }{3 t ^2+4 t +1}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{ t ^2+\frac{4 t }{3}+\frac{1}{3}}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{ t ^2+2\left(\frac{2 t }{3}\right)+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2+\frac{1}{3}}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{\left( t +\frac{2}{3}\right)^2+\left(\frac{-4+3}{9}\right)}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{\left( t +\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2}$
$=\frac{1}{3}\left[\frac{1}{2 \times \frac{1}{3}} \log \left|\frac{\left( t +\frac{2}{3}\right)-\frac{1}{3}}{\left( t +\frac{2}{3}\right)+\frac{1}{3}}\right|\right]_0^1$
$=\frac{1}{2}\left[\log \left|\frac{3 t +1}{3 t +3}\right|\right]_0^1 $
$=\frac{1}{2}\left[\log \left(\frac{4}{6}\right)-\log \left(\frac{1}{3}\right)\right]$
$=\frac{1}{2} \log \left(\frac{4}{6} \times 3\right)$
$\therefore I=\frac{1}{2} \log 2$
Put $\tan x=t$
$\therefore \sec ^2 x d x = dt$
When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$
$ \therefore I=\int_0^1 \frac{ dt }{3 t ^2+4 t +1}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{ t ^2+\frac{4 t }{3}+\frac{1}{3}}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{ t ^2+2\left(\frac{2 t }{3}\right)+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2+\frac{1}{3}}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{\left( t +\frac{2}{3}\right)^2+\left(\frac{-4+3}{9}\right)}$
$=\frac{1}{3} \int_0^1 \frac{ dt }{\left( t +\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2}$
$=\frac{1}{3}\left[\frac{1}{2 \times \frac{1}{3}} \log \left|\frac{\left( t +\frac{2}{3}\right)-\frac{1}{3}}{\left( t +\frac{2}{3}\right)+\frac{1}{3}}\right|\right]_0^1$
$=\frac{1}{2}\left[\log \left|\frac{3 t +1}{3 t +3}\right|\right]_0^1 $
$=\frac{1}{2}\left[\log \left(\frac{4}{6}\right)-\log \left(\frac{1}{3}\right)\right]$
$=\frac{1}{2} \log \left(\frac{4}{6} \times 3\right)$
$\therefore I=\frac{1}{2} \log 2$
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