b) Ta có
$\int\limits_{0}^{\pi/2}\frac{\sin 2x }{4-\cos ^{2}x}dx=\int\limits_{0}^{\pi/2}\frac{-2\sin 2x }{\cos 2x -7}dx=\int\limits_{0}^{\pi/2}\frac{(\cos 2x-7)'}{\cos 2x -7}dx=\left[ {\ln|\cos 2x -7|} \right]_{0}^{\pi/2}= \ln\dfrac{4}{3} $

a)
$\dfrac{1}{\tan x+1}=\dfrac{\cos x}{\sin x +\cos x}=\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{\cos x-\sin x}{\sin x +\cos x}=\dfrac{1}{2}(x)'+ \dfrac{1}{2}(\ln |\sin x+ \cos x|)'$
suy ra
$\int\limits_{0}^{\pi/4}\dfrac{1}{\tan x+1}dx=\dfrac{1}{2}\left[ {x+\ln |\sin x+ \cos x|} \right]_{0}^{\pi/4}=\dfrac{\pi+\ln 4}{8}$

b)
$ \int \tan^4x dx = \int(\frac{1}{\cos^2 x}\tan^2 x - \tan^2 x) dx = \frac{\tan^3 x}{3} + \int \frac{\cos^2 x - 1}{\cos^2 x} dx= \frac{\tan^3 x}{3} + x - \tan x +C$
Suy ra 
 $ \int\limits_{\pi/4}^{\pi/3} \tan^{4}xdx=\left[ { \frac{\tan^3 x}{3} + x - \tan x } \right]_{\pi/4}^{\pi/3}=\dfrac{\pi+8}{12}$

b)

b)
$I=\int\limits_{-\pi}^{\pi}\cos mx\cos nxdx$
 $=\int\limits_{-\pi}^{\pi} \frac{1}{2}(\cos(m+n)x+\cos(m-n)x)dx$
+ Nếu $m=n$
$I=\int\limits_{-\pi}^{\pi} \frac{1}{2}(\cos2mx+1)dx=\left[ { \frac{1}{4}\sin2mx+\frac{1}{2}x} \right]_{-\pi}^{\pi}=\pi$
+ Nếu $m=-n$
$I=\int\limits_{-\pi}^{\pi} \frac{1}{2}(\cos2mx+1)dx=\left[ { \frac{1}{4}\sin2mx+\frac{1}{2}x} \right]_{-\pi}^{\pi}=\pi$
+ Nếu $m\ne \pm n$
 $I=\frac{1}{2}(\frac{\sin(m+n)x}{m+n}+\frac{\sin(m-n)x}{m-n})dx\left|\begin{array}{l}\pi\\-\pi\end{array}\right.=\frac{\sin(m+n)\pi}{m+n}+\frac{\sin(m-n)\pi}{m-n}$