MCQ
$\int_0^{\frac{\pi}{2}} \log \sin x d x=$
- ✓$-\left(\frac{\pi}{2}\right) \log 2$
- B$\pi \log \frac{1}{2}$
- C$-\pi \log \frac{1}{2}$
- D$\frac{\pi}{2} \log 2$
✓
Answer
Correct option: A.
$-\left(\frac{\pi}{2}\right) \log 2$
(A)
Let $I =\int_0^{\frac{\pi}{2}} \log \sin x d x$
$=\int_0^{\frac{\pi}{4}}(\log \sin x+\log \cos x) d x$
$\ldots\left[\because \int_0^{2 a } f (x) d x=\int_0^{ a }[ f (x)+ f (2 a -x)] d x\right]$
$\begin{aligned}= & \int_0^{\frac{\pi}{4}} \log \sin x \cos x d x \\ = & \int_0^{\frac{\pi}{4}} \log \left(\frac{\sin 2 x}{2}\right) d x \\ = & \int_0^{\frac{\pi}{4}} \log \sin 2 x d x-\int_0^{\frac{\pi}{4}} \log 2 d x\end{aligned}$
In $1^{\text {st }}$ integral, $2 x= t \Rightarrow 2 d x= dt$
$\therefore \quad I=\frac{1}{2} \int_0^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{4} \log 2$
$=\frac{1}{2} \int_0^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{4} \log 2$
$\ldots\left[\because \int_{ b }^{ b } f (x) d x=\int_{ a }^{ b } f ( t ) dt \right]$
$\begin{array}{ll}\therefore & I=\frac{1}{2} I-\frac{\pi}{4} \log 2 \\ \therefore & I=\frac{-\pi}{2} \log 2\end{array}$
Let $I =\int_0^{\frac{\pi}{2}} \log \sin x d x$
$=\int_0^{\frac{\pi}{4}}(\log \sin x+\log \cos x) d x$
$\ldots\left[\because \int_0^{2 a } f (x) d x=\int_0^{ a }[ f (x)+ f (2 a -x)] d x\right]$
$\begin{aligned}= & \int_0^{\frac{\pi}{4}} \log \sin x \cos x d x \\ = & \int_0^{\frac{\pi}{4}} \log \left(\frac{\sin 2 x}{2}\right) d x \\ = & \int_0^{\frac{\pi}{4}} \log \sin 2 x d x-\int_0^{\frac{\pi}{4}} \log 2 d x\end{aligned}$
In $1^{\text {st }}$ integral, $2 x= t \Rightarrow 2 d x= dt$
$\therefore \quad I=\frac{1}{2} \int_0^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{4} \log 2$
$=\frac{1}{2} \int_0^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{4} \log 2$
$\ldots\left[\because \int_{ b }^{ b } f (x) d x=\int_{ a }^{ b } f ( t ) dt \right]$
$\begin{array}{ll}\therefore & I=\frac{1}{2} I-\frac{\pi}{4} \log 2 \\ \therefore & I=\frac{-\pi}{2} \log 2\end{array}$
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