A calculus problem by Vighnesh Raut

$\large f\left( x \right) =\int _{ \frac { { \pi }^{ 2 } }{ 16 } }^{ { x }^{ 2 } }{ \frac { \cos { x } \cos { \sqrt { \theta } } }{ 1+\sin ^{ 2 }{ \sqrt { \theta } } } d\theta}$

$\Rightarrow f^{ \prime }\left( x \right) =\frac { cos\quad x.cos\quad \sqrt { { x }^{ 2 } } }{ 1+{ sin }^{ 2 }\sqrt { { x }^{ 2 } } } .(2x)-\frac { cos\quad x.cos\quad \sqrt { \frac { { \pi }^{ 2 } }{ 16 } } }{ 1+{ sin }^{ 2 }\sqrt { \frac { { \pi }^{ 2 } }{ 16 } } } .(0)$

For knowing how the above step came, refer this .

$\Rightarrow f^{ \prime }\left( x \right) =\frac { cos\quad x.cos\quad x }{ 1+{ sin }^{ 2 }\quad x } .(2x)-0\quad =\quad \frac { 2x.{ cos }^{ 2 }\quad x }{ 1+{ sin }^{ 2 }\quad x }$

$Now, f^{ \prime }\left( \pi \right) =\frac { 2\pi .(1) }{ 1+0 } =\frac { 2\pi }{ 1 }$

$\Rightarrow \boxed{k = 1}$

$\begin{aligned} f(x) & = \int_{\frac {\pi^2}{16}}^{x^2} \frac {\cos x \cos \sqrt \theta}{1+\sin^2 \sqrt \theta} d\theta & \small \color{#3D99F6} \text{Let }t^2 = \theta \implies 2t \ dt = d\theta \\ & = \int_\frac \pi 4^x \frac {2t \cos x \cos t}{1+\sin^2 t} dt & \small \color{#3D99F6} \text{By fundamental theorem of calculus:} \\ \implies f'(x) & = \frac {2x \cos^2 x}{1+\sin^2 x} & \small \color{#3D99F6} S(x) = \int_a^x f(t) \ dt \implies S'(x) = f(x) \\ f'(\pi) & = 2\pi \end{aligned}$

Therefore, $k=\boxed 1$ .

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Reference: Fundamental theorems of calculus