If $I_1=\int_0^{\pi / 2} \frac{x}{\sin x} d x$, and $I_2=\int_0^1 \frac{ an ^{-1} x}{x} d x$, then $I_1: I_2$ is

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If $I_1=\int_0^{\pi / 2} \frac{x}{\sin x} d x$, and $I_2=\int_0^1 \frac{	an ^{-1} x}{x} d x$, then $I_1: I_2$ is

Aicharya · Accepted Answer

Given, $I_1=\int_0^{\pi / 2} \frac{x}{\sin x} d x$ $ I_2=\int_0^1 \frac{	an ^{-1} x}{x} d x=\int_0^{\pi / 4} \frac{t}{	an t} \cdot \sec ^2 t d t $ Put, $	an ^{-1} x=t$ $ \begin{aligned} & x=	an t \ & d x=\sec ^2 t d t \ & U . L ightarrow \frac{\pi}{4} \end{aligned} $ L. $L ightarrow 0$ $ \begin{aligned} & =\int_0^{\pi / 4} t \cdot \frac{\cos t}{\sin t} \cdot \frac{1}{\cos ^2 t} d t=\int_0^{\pi / 4} \frac{t}{\sin t \cdot \cos t} d t \ & =\int_0^{\pi / 4} \frac{2 t}{2 \sin t \cdot \cos t} d t=\int_0^{\pi / 4} \frac{2 t}{\sin 2 t} d t \end{aligned} $ $ \begin{aligned} & 	ext { Now, put } 2 t=x \ & 2 d t=d x \ & d t=\frac{d x}{2} \ & U . L ightarrow \frac{\pi}{2} \ & \end{aligned} $ $ \begin{aligned} & L . L ightarrow 0 \ & =2 \int_0^{\pi / 2} \frac{x}{\sin x} d x \ & I_2=2 \cdot I_1 \ & 	herefore \quad I_1: I_2=2: 1 \end{aligned} $ Hence, option (2) is correct

If $I_1=\int_0^{\pi / 2} \frac{x}{\sin x} d x$, and $I_2=\int_0^1 \frac{\tan ^{-1} x}{x} d x$, then $I_1:…

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