Chủ đề này có 1 trả lời

[inhtoan]

Tính $\int\limits_1^2 {{{{\rm{(arccot}}\sqrt {x - 1} )}^2}dx}$

[Crystal]

Tính $\int\limits_1^2 {{{{\rm{(arccot}}\sqrt {x - 1} )}^2}dx}$

Đặt $\left\{ \begin{array}{l}
u = arc{\cot ^2}\sqrt {x - 1} \\
dv = dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = - \frac{{arc\cot \sqrt {x - 1} }}{{x\sqrt {x - 1} }}dx\\
v = x
\end{array} \right.$
Khi đó: $$\int\limits_1^2 {{{\left( {arc\cot \sqrt {x - 1} } \right)}^2}dx} = \left. {xarc{{\cot }^2}\sqrt {x - 1} } \right|_1^2 + \int\limits_1^2 {\frac{{arc\cot \sqrt {x - 1} }}{{\sqrt {x - 1} }}dx} $$
$$ = - \frac{{{\pi ^2}}}{8} + \int\limits_1^2 {\frac{{arc\cot \sqrt {x - 1} }}{{\sqrt {x - 1} }}dx} $$
Lại đặt: $\left\{ \begin{array}{l}
u = arc\cot \sqrt {x - 1} \\
dv = \frac{{dx}}{{\sqrt {x - 1} }}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = - \frac{1}{{2x\sqrt {x - 1} }}dx\\
v = 2\sqrt {x - 1}
\end{array} \right.$
Suy ra: $$\int\limits_1^2 {\frac{{arc\cot \sqrt {x - 1} }}{{\sqrt {x - 1} }}dx} = \left. {2\sqrt {x - 1} arc\cot \sqrt {x - 1} } \right|_1^2 + \int\limits_1^2 {\frac{{dx}}{x}} $$
$$ = \frac{\pi }{2} + \left. {\ln \left| x \right|} \right|_1^2 = \frac{\pi }{2} + \ln 2$$
Do đó: $$\int\limits_1^2 {{{\left( {arc\cot \sqrt {x - 1} } \right)}^2}dx} = \boxed{ - \frac{{{\pi ^2}}}{8} + \frac{\pi }{2} + \ln 2}$$