Prove that for any positive integer $$k$$,
$${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$$
Hence prove that $$\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $$

If $$f$$ and $$g$$ are continuous function on $$\left[ {0,a} \right]$$ satisfying
$$f\left( x \right) = f\left( {a - x} \right)$$ and $$g\left( x \right) + g\left( {a - x} \right) = 2,$$
then show that $$\int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( x \right)dx} } $$

Evaluate $$\int\limits_0^1 {\log \left[ {\sqrt {1 - x} + \sqrt {1 + x} } \right]dx} $$

Evaluate : $$\int\limits_0^\pi {{{x\,dx} \over {1 + \cos \,\alpha \,\sin x}},0 < \alpha < \pi } $$