If int_0^pi frac{d x}{1+2 sin ^2 x}=k,then th | vedclass

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(C) Let $I = \int_0^\pi \frac{d x}{1+2 \sin ^2 x}$.Since the integrand $f(x) = \frac{1}{1+2 \sin ^2 x}$ satisfies $f(\pi - x) = f(x)$,we can write:$I

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(C) Let $I = \int_0^\pi \frac{d x}{1+2 \sin ^2 x}$.Since the integrand $f(x) = \frac{1}{1+2 \sin ^2 x}$ satisfies $f(\pi - x) = f(x)$,we can write:$I = 2 \int_0^{\pi / 2} \frac{d x}{1+2 \sin ^2 x}$.Dividing the numerator and denominator by $\cos ^2 x$:$I = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{\sec ^2 x + 2 	an ^2 x} d x = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{1 + 	an ^2 x + 2 	an ^2 x} d x = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{1 + 3 	an ^2 x} d x$.Let $	an x = t$,then $\sec ^2 x d x = d t$. When $x = 0, t = 0$ and when $x 	o \pi / 2, t 	o \infty$.$I = 2 \int_0^{\infty} \frac{d t}{1 + 3 t^2} = 2 \int_0^{\infty} \frac{d t}{1 + (\sqrt{3} t)^2}$.Using the formula $\int \frac{dx}{1+a^2x^2} = \frac{1}{a} 	an^{-1}(ax)$:$I = 2 \left[ \frac{1}{\sqrt{3}} 	an^{-1}(\sqrt{3} t) ight]_0^{\infty} = \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{\sqrt{3}}$.Given $k = \frac{\pi}{\sqrt{3}} \approx \frac{3.14159}{1.732} \approx 1.81$.The greatest integer less than or equal to $k$ is $\lfloor 1.81 floor = 1$.

If $\int_0^\pi \frac{d x}{1+2 \sin ^2 x}=k$,then the greatest integer less than or equal to $k$ is

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