Toán 12tính tích phân

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[imath]\displaystyle \int^{\pi / 2}_{0}{ \dfrac{ \cos{x}}{\sin^2{x}-5 \sin{x}+6}} \mathrm{d} x[/imath]
Đặt [imath]t = \sin{x} \ (t \in [-1;1]) \Rightarrow \mathrm{d} t = \cos{x} \mathrm{d} x [/imath]
Đổi cận:
[imath]x=0 \Rightarrow t=0 \\ x= \dfrac{ \pi}{2} \Rightarrow t=1[/imath]
Ta có: [imath]\displaystyle \int^{1}_{0}{ \dfrac{ \mathrm{d}t}{t^2-5t+6}}[/imath]

[imath]\dfrac{1}{t^2-5t+6} = \dfrac{1}{(t-2)(t-3)} \\ = \dfrac{A}{t-2} + \dfrac{B}{t-3} \\ = \dfrac{A(t-3) + B(t-2)}{(t-2)(t-3)} \\ = \dfrac{(A+B)t-3A-2B}{(t-2)(t-3)}[/imath]
Ta có hệ: [imath] \left\{\begin{matrix} A+B=0 \\ -3A-2B=1 \end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix} A=-1 \\ B=1 \end{matrix}\right. [/imath]

Nên ta có: [imath]\displaystyle \int^{1}_{0}{ \dfrac{ \mathrm{d}t}{t^2-5t+6}} = \displaystyle \int^{1}_{0}{ \left ( \dfrac{ -1}{t-2} + \dfrac{1}{t-3} \right )} \mathrm{d}t \\ =- \int^{1}_{0}{ \dfrac{ 1}{t-2} } \mathrm{d}t + \int^{1}_{0}{\dfrac{1}{t-3} } \mathrm{d}t \\ =- ( \ln|t-2|)|^1_0 + ( \ln|t-3|)|^1_0 \\ = -( \ln{1} - \ln{2} )+ ( \ln{2} - \ln{3}) \\ = 2 \ln{2} - \ln{3} \\ = \ln{2^2} - \ln{3} \\ = \ln{ \left ( \dfrac{4}{3} \right )}[/imath]

[imath]\displaystyle \int^{\pi / 2}_{0}{ \dfrac{ \cos{x}}{\sin^2{x}-5 \sin{x}+6}} \mathrm{d} x = a \ln { \left ( \dfrac{4}{c} \right )} +b[/imath]
Suy ra [imath]a=1, \ b=0, \ c=3[/imath]
[imath]S=a+b+c=4[/imath]

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