Evaluation of Definite Integral-Different Methods

Find ${\int }_0^{\infty }\text{f}\left(\frac{\text{a}}{\text{x}}+\frac{\text{x}}{\text{a}}\right)·\frac{\ell \text{nx}}{\text{x}}\text{dx}$

The value of ${\int }_0^{\infty }\frac{dx}{x^2+2x \cos \theta +1}$ is 

$I={\int }_0^{\pi }\frac{x^{2 }\sin 2x\sin \left(\frac{\pi }{2}\cos x\right)}{2x-\pi } dx$

For $\text{f}\left(\text{x}\right)={\text{x}}^4+\left|\text{x}\right|$, let ${\text{I}}_1={\int }_0^{\pi }\text{f}\left(\text{cos x}\right)\text{ dx}$ and ${\text{I}}_2={\int }_0^{\pi /2}\text{f}\left(\text{sin x}\right)\text{ dx }$then $\frac{{\text{I}}_1}{{\text{I}}_2}$ has the value of equal to

${\int }_0^{\infty }\frac{dx}{\left(1+x^a\right)\left(1+x^2\right)}; \left(a>0\right)$

The value of : $\underset{0}{\overset{2}{\int }}\sqrt{4-x^2}dx$ would be:

${\int }_0^{\frac{\pi }{4}}\log \left(1+\tan x\right)dx=$

If $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$, then the value of $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{1+\tan x}$ is

The value of ${\int }_1^{e^{37}}\frac{\mathrm{\pi sin}\left(\mathrm{\pi }\ell nx\right)}{x}dx$ is

The value of  ${\int }_1^{e^{37}}\frac{\pi \sin \left(\pi \ln x\right)}{x}dx$  is:

The value of  ${\displaystyle {\int }_2^3\frac{\sqrt{x}}{\sqrt{\mathrm{}\mathrm{}\mathrm{}5-x}+\sqrt{x}}dx}$  is:

The integral  ${\int }_0^{1.5}\left[x^2\right]dx,$ where $\left[.\right]$ denotes the greatest integer function, equals:

If $I\left(m,n\right)=\underset{0}{\overset{1}{\int }}{t}^m{\left(1+t\right)}^{n}dt$, then the expression for $I\left(m,n\right)$ in terms of $I\left(m+1,n-1\right)$ where $m, n\in N$ is

Let $f$ be a positive function.

Let $I_1=\underset{1-k}{\overset{k}{\int }}xf\left[x\left(1-x\right)\right]dx, I_2=\underset{1-k}{\overset{k}{\int }}f\left[x\left(1-x\right)\right]dx,$ where $2k-1>0.$ Then $\frac{I_1}{I_2}$ is