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

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Question

$f (x) =$ $\int\limits_0^\pi  {\frac{{t\,\sin t}}{{\sqrt {1 + {{	an }^2}x\,{{\sin }^2}t} }}} \,dt$

Using king and add.

$f (x) = \frac{\

Vedclass · Accepted Answer

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

Vedclass · Answer

$f (x) =$ $\int\limits_0^\pi  {\frac{{t\,\sin t}}{{\sqrt {1 + {{	an }^2}x\,{{\sin }^2}t} }}} \,dt$

Using king and add.

$f (x) = \frac{\pi }{2}\,\,\int\limits_0^\pi  {\frac{{\,\sin t}}{{\sqrt {1 + {{	an }^2}x\,{{\sin }^2}t} }}} \,dt$  $= \pi \,\int\limits_0^{\frac{\pi }{2}} {\frac{{\,\sin t}}{{\sqrt {1 + {{	an }^2}x\,(1 - {{\cos }^2}t)} }}} \,dt$

$= \pi \,\int\limits_0^{\frac{\pi }{2}} {\frac{{\,\sin t}}{{\sqrt {{{\sec }^2}x\, - \,{{	an }^2}x\,{{\cos }^2}t} }}} \,dt$ $= \pi \,\int\limits_0^1 {\frac{{dy}}{{\sqrt {{{\sec }^2}x\, - \,{{	an }^2}x\,.{y^2}\,} }}} \,$

$= \frac{\pi }{{	an x}}\,\int\limits_0^1 {\frac{{dy}}{{\sqrt {\cos e{c^2}x\, - \,\,{y^2}\,} }}} \,$  $= \frac{\pi }{{	an x}}\,\left\{ {{{\sin }^{ - 1}}\frac{y}{{\cos ecx}}} ight\}_0^1\,\, = \,\,\frac{\pi }{{	an x}}\,{\sin ^{ - 1}}(\sin x)\, = \,\frac{{\pi \,x}}{{	an x}}\,$

If $f (x) = \int\limits_0^\pi {\frac{{t\,\sin t\,dt}}{{\sqrt {1 + {{\tan }^2}x\,\,\,{{\sin }^2}t} }}\,} $ for $0 < x < \frac{\pi }{2}\,$

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