Mnguoi giúp em với ạ.

Đặt 2x=t dx=$\frac{dt}{2}$ 

I=$\int\limits{\frac{1}{4}\frac{t}{(sint)^2}} \, dt$

Đặt $\left \{ {{u=t} \atop {dv=\frac{dt}{[sin(t)]^{2}}}} \right.$ =>$\left \{ {{du=dt} \atop {v=-cot(t)}} \right.$ .

Áp dụng công thức uv$-$$\int\limits {v} \, du$ =-t.cot(t) $-$$\int\limits {-cot(t)} \, dt$ = $-$t.cot(t) $+$ln|sin(t)|+C

Vậy I=$\frac{1}{4}$.(-2xcot(2x)+ln|sin(2x|)+C=$\frac{-x}{2}$.cot(2x) + $\frac{1}{4}$.ln|sin2x|+C

$\begin{array}{l}
b)I = \int {\dfrac{x}{{{{\sin }^2}2x}}dx} \\
\left\{ \begin{array}{l}
x = u\\
\dfrac{1}{{{{\sin }^2}2x}}dx = dv
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
dx = du\\
 - \dfrac{{\cot 2x}}{2} = v
\end{array} \right.\\
 \Rightarrow I =  - \dfrac{{x\cot 2x}}{2} + \dfrac{1}{2}\int {\cot 2xdx} \\
 =  - \dfrac{{x\cot 2x}}{2} + \dfrac{1}{2}\int {\dfrac{{\cos 2x}}{{\sin 2x}}dx} \\
 =  - \dfrac{{x\cot 2x}}{2} + \dfrac{1}{4}\int {\dfrac{{2\cos 2x}}{{\sin 2x}}dx} \\
 =  - \dfrac{{x\cot 2x}}{2} + \dfrac{1}{4}\int {\dfrac{{d\left( {\sin 2x} \right)}}{{\sin 2x}}} \\
 =  - \dfrac{{x\cot 2x}}{2} + \dfrac{1}{4}\ln \left| {\sin 2x} \right| + C
\end{array}$