Questions in definite-integral

Select	Question
	$\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+.....\frac{1}{2n} \right]=$
	$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\frac{k}{{{n}^{2}}+{{k}^{2}}}}$is equals to
	$\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ \frac{1}{n}+\frac{1}{\sqrt{{{n}^{2}}+n}}+\frac{1}{\sqrt{{{n}^{2}}+2n}}+.....+\frac{1}{\sqrt{{{n}^{2}}+(n-1)n}} \right]$ is equal to
	$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{p}}+{{2}^{p}}+{{3}^{p}}+.....+{{n}^{p}}}{{{n}^{p+1}}}=$
	$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\frac{1}{n}{{e}^{\frac{r}{n}}}}$is
	$\int_{0}^{\infty }{\frac{\log \,(1+{{x}^{2}})}{1+{{x}^{2}}}}\,dx=$
	$\int_{0}^{\pi /2}{{{\sin }^{2}}x{{\cos }^{3}}x}\,dx=$
	$\int_{-\pi /2}^{\pi /2}{{{\sin }^{2}}x\,dx=}$
	The correct evaluation of $\int_{0}^{\pi }{\left\| \,{{\sin }^{4}}x\, \right\|\,dx}$ is
	$\int_{0}^{\pi /2}{{{\cos }^{2}}x\,dx=}$