Q114 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
$\int \frac{\left(1-4 \sin ^2 x\right) \cos x}{\cos (3 x+2)} d x=$
- A$(\cos 2) x-\frac{1}{3}(\sin 2) \log |\sec (3 x+2)|+c$
- B$(\sin 2) x-\frac{1}{3}(\cos 2) \log |\cos (3 x+2)|+c$
- C$(\sin 2) x+\frac{1}{3}(\cos 2) \log |\cos (3 x+2)|+c$
- Correct answerD$(\cos 2) x+\frac{1}{3}(\sin 2) \log |\sec (3 x+2)|+c$
Solution
$\int \frac{\left(1-4 \sin ^2 x\right) \cos \pi}{\cos (3 x+2)} d x=\int \frac{\left[1-4\left(1-\cos ^2 x\right)\right] \cos x}{\cos (3 \pi+2)} d x$
$\begin{aligned} & =\int \frac{4 \cos { }^3 x-3 \cos x}{\cos (3 x+2)} d x=\int \frac{\cos 3 x}{\cos (3 x+2)} d x \\ & \Rightarrow \operatorname{Let} 3 x+2=t \Rightarrow d t=3 c d x \\ & \mathbb{I}=\int \frac{\cos (t-2)}{\cos t} \frac{d t}{3}=\frac{1}{3} \int \frac{(\cos t \cos 2+\sin t \sin 2)}{\cos t} d t \\ & =\frac{1}{3} \int(\cos 2+\sin 2 \tan t) d t=\frac{1}{3}\{t \cos 2+\sin 2 \ln (\sec t)\}+c \\ & =x \cos 2+\frac{1}{3} \sin 2 \ln |\sec (3 x+2)|+c\end{aligned}$
Hence, the correct answer is option: 'D'.