Q112 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
$\int \frac{\sec x}{3(\sec x+\tan x)+2} d x=$
- Correct answerA$\frac{1}{2} \log \left|\frac{\tan \frac{x}{2}+1}{\tan \frac{x}{2}+5}\right|+c$
- B$\frac{2}{\sqrt{11}} \tan ^{-1}\left(\frac{3 \tan \frac{x}{2}+4}{\sqrt{11}}\right)+c$
- C$\log |3 \sec x+2 \tan x|+c$
- D$\log |3 \tan x+2 \sec x|+c$
Solution
$\begin{aligned} & I=\int \frac{\sec x(\sec x-\tan x) d x}{3\left(\sec ^2 x-\tan ^2 x\right)+2(\sec x-\tan x)} \\ & \sec x-\tan x=t \\ & I=\int \frac{-d t}{3+2 t}=\frac{-1}{2} \ln (3+2 t)+C \\ &= \frac{-1}{2} \ln (3+2 \sec x-2 \tan x)+C \\ &= \frac{-1}{2} \ln \left(\frac{3 \cos x+2-2 \sin x}{\cos x}\right)+C \\ &= \frac{-1}{2} \ln \left[\frac{3-3 \tan ^2 \frac{x}{2}+2\left(1+\tan ^2 \frac{x}{2}\right)-4 \tan \frac{x}{2}}{1-\tan ^2 \frac{x}{2}}\right]+C \\ &= \frac{-1}{2} \ln \left[\frac{5-4 \tan ^2 \frac{x}{2}-\tan ^2 \frac{x}{2}}{1-\tan ^2 \frac{x}{2}}\right]+C\end{aligned}$
$\begin{aligned} & =\frac{1}{2} \ln \left[\frac{1-\tan ^2 \frac{x}{2}}{\left(1-\tan \frac{x}{2}\right)\left(5+\tan \frac{x}{2}\right)}\right]+C \\ & =\frac{1}{2} \ln \left(\frac{1+\tan \frac{x}{2}}{5+\tan \frac{x}{2}}\right)+C .\end{aligned}$
Hence, the correct answer is option: 'A'.