If f(x) = intlimits_0^pi {frac{t sin t , dt}{ | vedclass

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(C) Given $f(x) = \int\limits_0^\pi \frac{t \sin t}{\sqrt{1 + 	an^2 x \sin^2 t}} \, dt$. Using the property $\int_0^a f(t) \, dt = \int_0^a f(a-t) \,

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$f$ is continuous and differentiable in $\left(0, \frac{\pi}{2}\right)$

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(C) Given $f(x) = \int\limits_0^\pi \frac{t \sin t}{\sqrt{1 + 	an^2 x \sin^2 t}} \, dt$. Using the property $\int_0^a f(t) \, dt = \int_0^a f(a-t) \, dt$,we have $f(x) = \int_0^\pi \frac{(\pi-t) \sin(\pi-t)}{\sqrt{1 + 	an^2 x \sin^2(\pi-t)}} \, dt = \int_0^\pi \frac{(\pi-t) \sin t}{\sqrt{1 + 	an^2 x \sin^2 t}} \, dt$. Adding the two expressions: $2f(x) = \pi \int_0^\pi \frac{\sin t}{\sqrt{1 + 	an^2 x \sin^2 t}} \, dt$. Since the integrand is symmetric about $t = \frac{\pi}{2}$,$f(x) = \pi \int_0^{\pi/2} \frac{\sin t}{\sqrt{1 + 	an^2 x \sin^2 t}} \, dt$. Let $y = \cos t$,then $dy = -\sin t \, dt$. When $t=0, y=1$; when $t=\pi/2, y=0$. $f(x) = \pi \int_0^1 \frac{dy}{\sqrt{1 + 	an^2 x (1-y^2)}} = \pi \int_0^1 \frac{dy}{\sqrt{\sec^2 x - 	an^2 x \, y^2}} = \frac{\pi}{	an x} \int_0^1 \frac{dy}{\sqrt{\csc^2 x - y^2}}$. $f(x) = \frac{\pi}{	an x} [\sin^{-1}(\frac{y}{\csc x})]_0^1 = \frac{\pi}{	an x} \sin^{-1}(\sin x) = \frac{\pi x}{	an x}$. Since $f(x) = \frac{\pi x}{	an x}$ is a composition of differentiable functions in $(0, \pi/2)$,$f$ is continuous and differentiable in $(0, \pi/2)$.

If $f(x) = \int\limits_0^\pi {\frac{t \sin t \, dt}{\sqrt{1 + \tan^2 x \sin^2 t}}}$ for $0 < x < \frac{\pi}{2}$,then which of the following is true?

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