Đặt $I=\int\limits_0^{\pi/2}\frac{\cos^{2014} x}{\sin^{2014} x+\cos^{2014} x}dx,J=\int\limits_0^{\pi/2}\frac{\sin^{2014} x}{\sin^{2014} x+\cos ^{2014}x}dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\frac{\cos^{2014}(\displaystyle\frac{\pi}{2}-t)}{\sin^{2014}(\displaystyle\frac{\pi}{2}-t)+\cos^{2014}(\displaystyle\frac{\pi}{2}-t)}dt$
   $=\int\limits_0^{\pi/2}\frac{\sin^{2014} t}{\sin^{2014} t+\cos^{2014} t}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\frac{\pi}{2}$
Suy ra: $I=J=\frac{\pi}{4}$

Đặt $I=\int\limits_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx,J=\int\limits_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\frac{\sqrt{\sin(\displaystyle\frac{\pi}{2}-t)}}{\sqrt{\sin(\displaystyle\frac{\pi}{2}-t)}+\sqrt{\cos(\displaystyle\frac{\pi}{2}-t)}}dt$
   $=\int\limits_0^{\pi/2}\frac{\sqrt{\cos t}}{\sqrt{\sin t}+\sqrt{\cos t}}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\frac{\pi}{2}$
Suy ra: $I=J=\frac{\pi}{4}$

Đặt $I=\int\limits_0^{\pi/2}\frac{\cos^nx}{\cos^n x+\sin^n x}dx,J=\int\limits_0^{\pi/2}\frac{\sin^n x}{\sin^n x+\cos^n x}dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\frac{\cos^n(\displaystyle\frac{\pi}{2}-t)}{\sin^n(\displaystyle\frac{\pi}{2}-t)+\cos^n(\displaystyle\frac{\pi}{2}-t)}dt$
   $=\int\limits_0^{\pi/2}\frac{\sin^nt}{\sin^n t+\cos^n t}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\frac{\pi}{2}$
Suy ra: $I=J=\frac{\pi}{4}$

Đặt: $I=\int\limits_{-1/2}^{1/2}\sin\left(\ln\frac{1-x}{1+x}\right)dx$
Đặt: $x=-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{1/2}^{-1/2}\sin\left(\ln\frac{1+t}{1-t}\right)dt$
   $=\int\limits_{-1/2}^{1/2}\sin\left(-\ln\frac{1-t}{1+t}\right)dt$
   $=-\int\limits_{-1/2}^{1/2}\sin\left(\ln\frac{1-t}{1+t}\right)dt=-I$
Suy ra: $2I=0\Leftrightarrow I=0$

Đặt: $I=\int\limits_{-1/2}^{1/2}\cos x\ln\frac{1-x}{1+x}dx$
Đặt: $x=-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{1/2}^{-1/2}\cos(-t)\ln\frac{1+t}{1-t}dt$
   $=\int\limits_{-1/2}^{1/2}\cos t\ln\frac{1+t}{1-t}dt$
   $=-\int\limits_{-1/2}^{1/2}\cos t\ln\frac{1-t}{1+t}dt=-I$
Suy ra: $2I=0\Leftrightarrow I=0$

Ta có:
     $\int\limits_{-1}^{1}\frac{x^{4} + x^{2}}{x^{2}+1}dx$
$=\int\limits_{-1}^{1}x^2dx$
$=\frac{x^3}{3}\left|\begin{array}{l}1\\-1\end{array}\right.=\frac{2}{3}$

Đặt $I=\int\limits_0^{\pi/2}(\sqrt{\sin x}-\sqrt{\cos x})dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\left(\sqrt{\sin(\displaystyle\frac{\pi}{2}-t)}-\sqrt{\cos(\displaystyle\frac{\pi}{2}-t)}\right)dt$
   $=\int\limits_0^{\pi/2}(\sqrt{\cos t}-\sqrt{\sin t})dt=-I$
Suy ra: $2I=0\Leftrightarrow I=0$

Đặt $I=\int\limits_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx,J=\int\limits_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\frac{\sin(\displaystyle\frac{\pi}{2}-t)}{\sin(\displaystyle\frac{\pi}{2}-t)+\cos(\displaystyle\frac{\pi}{2}-t)}dt$
   $=\int\limits_0^{\pi/2}\frac{\cos t}{\sin t+\cos t}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\frac{\pi}{2}$
Suy ra: $I=J=\frac{\pi}{4}$

Đặt $I=\int\limits_0^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx,J=\int\limits_0^{\pi/2}\frac{\cos^3x}{\sin^3x+\cos^3x}dx$
Đặt: $x=\frac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\frac{\sin^3(\displaystyle\frac{\pi}{2}-t)}{\sin^3(\displaystyle\frac{\pi}{2}-t)+\cos^3(\displaystyle\frac{\pi}{2}-t)}dt$
   $=\int\limits_0^{\pi/2}\frac{\cos^3t}{\sin^3t+\cos^3t}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\frac{\pi}{2}$
Suy ra: $I=J=\frac{\pi}{4}$

Ta có:
$P=\frac{a}{4}+\frac{1}{a}+\frac{3a}{4}$
     $\ge2\sqrt{\frac{a}{4}.\frac{1}{a}}+\frac{3.2}{4}=\frac{5}{2}$
Min$P=\frac{5}{2}\Leftrightarrow a=2$

Có $6$ cách chọn vị trí cho chữ số $0$
Có $C_6^3$ cách chọn vị trí cho chữ số $4$
 Có $3$ cách chọn vị trí cho chữ số $1$
 Có $2$ cách chọn vị trí cho chữ số $2$
 Có $1$ cách chọn vị trí cho chữ số $3$
 Suy ra có: $6.C_6^3.3.2.1=720$ số thỏa mãn.

Ta có: $\begin{cases}x-2y= 4-m\\ 2x+y=3m+3 \end{cases}$
$\Leftrightarrow \begin{cases}x-2y= 4-m\\ 4x+2y=6m+6 \end{cases}$
$\Leftrightarrow \begin{cases}x=m+2\\y=m-1 \end{cases}$
Suy ra:
$A=x^2+y^2$
     $=(m+2)^2+(m-1)^2$
     $=2m^2+2m+5$
     $=2(m+\frac{1}{2})^2+\frac{9}{2}\ge\frac{9}{2}$
Min$A=\frac{9}{2}\Leftrightarrow m=\frac{-1}{2}$

Phương trình tương đương với:
      $3-4\sin^22x=2\cos2x(1+2\sin x)$
$\Leftrightarrow 1+2(1-2\sin^22x)-2\cos 2x-4\cos 2x\sin x=0$
$\Leftrightarrow 1+2(\cos4x-\cos2x)-2(\sin3x-\sin x)=0$
$\Leftrightarrow 1-4\sin3x\sin x-2\sin 3x+2\sin x=0$
$\Leftrightarrow (1+2\sin x)(1-2\sin3x)=0$
$\Leftrightarrow \left[ \begin{array}{l} \sin x=\frac{-1}{2}\\\sin3x=\frac{1}{2} \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{18}+k\frac{2\pi}{3}\\x=\frac{5\pi}{18}+k\frac{2\pi}{3} \end{array} \right.,k\in\mathbb{Z}$