Questions in definite-integral

Select	Question
	$\int_{0}^{\pi /4}{\frac{dx}{{{\cos }^{4}}x-{{\cos }^{2}}x{{\sin }^{2}}x+{{\sin }^{4}}x}=}$
	The value of $\int_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}\,dt+\int_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}\,dt}}$ is
	If for non-zero $x,$ $af(x)+bf\left( \frac{1}{x} \right)=\frac{1}{x}-5,$ where $a\ne b,$ then $\int_{1}^{2}{f(x)\,dx=}$
	If ${{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}\theta \,d\theta ,}$ then ${{I}_{8}}+{{I}_{6}}$ equals
	$\int_{0}^{\pi /4}{{}}\sec x\log (\sec x+\tan x)\,dx=$
	$\int_{0}^{\pi /4}{\frac{{{\sec }^{2}}x}{(1+\tan x)(2+\tan x)}}\,dx=$
	$\int_{0}^{2/3}{\frac{dx}{4+9{{x}^{2}}}=}$
	The value of $\int_{0}^{1}{\frac{{{x}^{4}}+1}{{{x}^{2}}+1}\,dx}$ is
	$\int_{0}^{a}{{{x}^{2}}\sin {{x}^{3}}\,dx}$ equals
	$\int_{0}^{\pi /4}{[\sqrt{\tan x}+\sqrt{\cot x}]\,dx}$ equals