The integral $\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$ is equal to (where c is a constant of integration)
Let f : R $\to$ R be a differentiable function such that $f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$ and $f'\left( {{\pi \over 2}} \right) = 1$ and let $g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt} $ for $x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$. Then $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$ is equal to :
Let f : R $\to$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $\in$ R where k > 0 and n is a positive integer. If ${I_1} = \int\limits_0^{4nk} {f(x)dx} $ and ${I_2} = \int\limits_{ - k}^{3k} {f(x)dx} $, then :
Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } $, $0 \le x \le 1$ and f(0) = 0, then $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} $ :
Let $f(x)=x+\dfrac{a}{\pi^{2}-4}\sin x+\dfrac{b}{\pi^{2}-4}\cos x,\ x\in\mathbb{R}$ be a function which satisfies
$\displaystyle f(x)=x+\int_{0}^{\pi/2}\sin(x+y)\,f(y)\,dy.$
Then $(a+b)$ is equal to:
The value of $\dfrac{e^{-\pi/4}+\displaystyle\int_{0}^{\pi/4} e^{-x}\tan^{50}x\,dx}{\displaystyle\int_{0}^{\pi/4} e^{-x}\big(\tan^{49}x+\tan^{51}x\big)\,dx}$ is: