Q116 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$\int_{\frac{-3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) d x=$

  • A
    $-\frac{\pi}{2} \log 2$
  • Correct answer
    B
    $-\frac{\pi}{8} \log 2$
  • C
    $-\frac{\pi}{14} \log 2$
  • D
    $-\frac{\pi}{28} \log 2$

Solution

Put $4 x+3=t \Rightarrow d x=\frac{a t}{4}$ $\begin{aligned} & x=\frac{\pi-6}{8} \Rightarrow t=\frac{\pi}{2} \\ & x=\frac{-3}{4} \Rightarrow t=0 \\ & \therefore I=\int_0^{\frac{\pi}{2}} \log (\sin t) \frac{d t}{4}=\frac{1}{4} \times\left(\frac{-\pi}{2} \log 2\right)=-\frac{\pi}{8} \log 2 .\end{aligned}$

Hence, the correct answer is option: 'B'.

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