If $\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=A \sin 2 x+B$,then $A$ is equal to

$\int \frac{d x}{\sin x+\cos x}=$

If $\int \frac{1}{x^4+8 x^2+9} d x = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right] + c$, then $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} =$

$\int {\frac{{2x + 5}}{{\sqrt {7 - 6x - {x^2}} }}dx} = A\sqrt {7 - 6x - {x^2}} + B\,{\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + C$ (where $C$ is a constant of integration),then the ordered pair $(A, B)$ is equal to

$\int {\left( {\sin \left( {101x} \right).{{\sin }^{99}}x} \right)} dx = \frac{{\sin \left( {100x} \right){{\left( {\sin x} \right)}^\lambda }}}{\mu } + C$ where $C$ is the constant of integration,then $\frac{\lambda }{\mu }$ is equal to

$\int \frac{x+\sin x}{1+\cos x} d x$ is equal to