Integrals Test ...

Evaluate the integral
$$\displaystyle \int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$$

$$\displaystyle \int_{0}^{a}\frac{1}{a^{2}+x^{2}}dx_{=}$$

Evaluate the integral
$$\displaystyle \int_{\frac{a}{2}}^{a}\frac{1}{\sqrt{a^{2}-x^{2}}}dx$$

Evaluate the integral
$$\displaystyle \int_{0}^{1}\frac{x}{1+x^{2}}dx$$

The integral $$\displaystyle \int_{0}^{1}\frac{x^{3}}{1+x^{8}}dx=$$

$$\displaystyle \int_{0}^{1}\frac{1}{1+x}dx_{=}$$

The integral $$\displaystyle \int_{0}^{1}\frac{(\mathrm{t}\mathrm{a}\mathrm{n}^{-1}{\mathrm{x})^{3}}}{1+\mathrm{x}^{2}}\mathrm{d}\mathrm{x}=$$

$$\displaystyle \int_{0}^{1/2}\mathrm{e}^{\mathrm{x}}\left[ { s }{ i }{ n }^{ -1 }{ x }+\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  }  \right] $$ dx $$=$$

The integral $$\displaystyle \int_{0}^{\pi/4} \displaystyle \frac{e^{\tan x}}{\cos^{2}x}dx=$$

If  $$\displaystyle \int_{0}^{k}\frac{\cos x}{1+\sin^{2}x}dx=\frac{\pi}{4}$$ then $${k}=?$$