Q111 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
If $\int \log \left(6 \sin ^2 x+17 \sin x+12\right)^{\cos x} d x=f(x)+c$ then, $f\left(\frac{\pi}{2}\right)=$
- A$\frac{1}{6}\left[\log 5^5+\log 7^7-12\right]$
- B$\frac{1}{6}[7 \log 5+5 \log 7+29]$
- C$\frac{1}{6}[14 \log 5+15 \log 7+12]$
- Correct answerD$\frac{1}{6}[15 \log 5+14 \log 7-29]$
Solution
$\mathrm{I}=\int \log \left(6 \sin ^2 x+17 \sin x+12\right)^{\cos x} d x$
Let $\sin x=t \Rightarrow \cos x d x=d t$ $\begin{aligned} & \mathrm{I}=\int \log \left(6 t^2+17 t+12\right) d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-\int \frac{t(12 t+17)}{6 t^2+17 t+12} d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-\int 2 d t+\int \frac{17 t+24}{6 t^2+17 t+12} d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-2 t+\frac{17}{12} \int \frac{12 t+17}{6 t^2+17 t+12} d t \\ & \quad-\frac{1}{12} \int \frac{1}{6 t^2+17 t+12} d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-2 t+\frac{17}{12} \log \left(6 t^2+17 t+12\right) \\ & -\frac{1}{4} \int \frac{d t}{3 t+4}+\frac{1}{6} \int \frac{d t}{2 t+3} \end{aligned}$ $\begin{aligned} & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-2 t+\frac{17}{12} \log \left(6 t^2+17 t+12\right) \\ & -\frac{\log (3 t+4)}{12}+\frac{\log (2 t+3)}{12}+\mathrm{C} \\ & \Rightarrow \mathrm{I}=\frac{\left[\begin{array}{c} (12 \sin x+17) \log ((2 \sin x+3)(3 \sin x+4)] \\ -\log (3 \sin x+4)+\log (2 \sin x+3)-24 \sin x \end{array}\right]}{12}+\mathrm{c}-\frac{17}{6}\left\{\because C=c-\frac{17}{6}\right\} \\ & \Rightarrow \mathrm{f}\left(\frac{\pi}{2}\right)=\frac{29 \log [35]-\log 7+\log 5-24+28 \log 7-28 \log 7}{12}-\frac{17}{6} \\ & =\frac{29 \log \frac{35}{7}+28 \log 7+\log 5-24}{12}-\frac{17}{6}=\frac{30 \log 5+28 \log 7-24}{12}-\frac{17}{6} \\ & =\frac{1}{6}[15 \log 5+14 \log 7-29] \end{aligned}$
Let $\sin x=t \Rightarrow \cos x d x=d t$ $\begin{aligned} & \mathrm{I}=\int \log \left(6 t^2+17 t+12\right) d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-\int \frac{t(12 t+17)}{6 t^2+17 t+12} d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-\int 2 d t+\int \frac{17 t+24}{6 t^2+17 t+12} d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-2 t+\frac{17}{12} \int \frac{12 t+17}{6 t^2+17 t+12} d t \\ & \quad-\frac{1}{12} \int \frac{1}{6 t^2+17 t+12} d t \\ & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-2 t+\frac{17}{12} \log \left(6 t^2+17 t+12\right) \\ & -\frac{1}{4} \int \frac{d t}{3 t+4}+\frac{1}{6} \int \frac{d t}{2 t+3} \end{aligned}$ $\begin{aligned} & \mathrm{I}=t \log \left(6 t^2+17 t+12\right)-2 t+\frac{17}{12} \log \left(6 t^2+17 t+12\right) \\ & -\frac{\log (3 t+4)}{12}+\frac{\log (2 t+3)}{12}+\mathrm{C} \\ & \Rightarrow \mathrm{I}=\frac{\left[\begin{array}{c} (12 \sin x+17) \log ((2 \sin x+3)(3 \sin x+4)] \\ -\log (3 \sin x+4)+\log (2 \sin x+3)-24 \sin x \end{array}\right]}{12}+\mathrm{c}-\frac{17}{6}\left\{\because C=c-\frac{17}{6}\right\} \\ & \Rightarrow \mathrm{f}\left(\frac{\pi}{2}\right)=\frac{29 \log [35]-\log 7+\log 5-24+28 \log 7-28 \log 7}{12}-\frac{17}{6} \\ & =\frac{29 \log \frac{35}{7}+28 \log 7+\log 5-24}{12}-\frac{17}{6}=\frac{30 \log 5+28 \log 7-24}{12}-\frac{17}{6} \\ & =\frac{1}{6}[15 \log 5+14 \log 7-29] \end{aligned}$
Hence, the correct answer is option: 'D'.