Properties of Definite Integrals

If $f\left(x\right)={\int }_0^x{\sin }^{6}tdt$, then $f\left(x+\mathrm{\pi }\right)=$

${\int }_0^{\sqrt{\log \frac{\mathrm{\pi }}{2}}}\cos \left(e^{x^2}\right)2x{e}^{x^2}dx=$

The equation ${\int }_{\frac{-\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\left\{a\left|\sin x\right|+\frac{b\sin x}{1+\cos x}+c\right\}dx=0$, where $a, b, c$ are constants, gives a relation between

${\int }_0^{2\mathrm{\pi }}\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx, n>0$ is equal to

${\int }_{-1}^1\sqrt{\frac{1-x}{1+x}}dx=$

The value of the integral $I={\int }_0^{1}x{\left(1-x\right)}^{n}dx$ is

If $I_1={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx, I_2={\int }_0^{2\mathrm{\pi }}{\cos }^{6}xdx$, ${\mathit{I}}_3={\int }_{\frac{-\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\sin }^{3}xdx, I_4={\int }_0^1\log \left(\frac{1-x}{x}\right)dx$, then

${\int }_0^{100\mathrm{\pi }}\left(\left|{\sin }^{3}x\right|+\left|{\cos }^{3}x\right|\right)dx=$

Let $f\left(x\right)$ be a function satisfying $f^{\text{'}}\left(x\right)=f\left(x\right)$ with $f\left(0\right)=1$ and $g\left(x\right)$ be the function satisfying $f\left(x\right)+g\left(x\right)=x^2$. The value of integral ${\int }_0^{1}f\left(x\right)g\left(x\right)dx$ is equal to

${\int }_0^{\frac{\pi }{2}}{\cos }^{6}xdx=$

${\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{1+\cos 2x}$ is equal to

The value of ${\int }_{-1}^1\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right)dx$ is

${\int }_{\frac{-\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{1+{\cos }^{2}x}{e}^{-{\cos }^{2}x}dx$ is equal to

${\int }_{-a}^a\sin xf\left(\cos x\right)dx=$

The value of ${\int }_{-a}^a\frac{1}{x+x^3}dx$ is

The value of $I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left|\sin x\right|dx$ is

${\int }_{-1}^{1}x\left|x\right|dx=$

${\int }_0^{\mathrm{\pi }}\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx=$

${\int }_0^{\mathrm{\pi }}{\mathrm{logsin}}^{2}xdx=$

The value of ${\int }_0^{2\mathrm{\pi }}{\cos }^{99}xdx$ is