xét sự hội tụ của tích phân suy rộng loại 2 giúp e với ạaaa

Đáp án:

Tích phân phân kỳ

Giải thích các bước giải:

\(\begin{array}{l}
\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\dfrac{1}{\sqrt{\sin x}.\cos x}dx = \displaystyle\int\limits_0^{1}\dfrac{1}{\sqrt{\sin x}.\cos x}dx+\displaystyle\int\limits_1^{\tfrac{\pi}{2}}\dfrac{1}{\sqrt{\sin x}.\cos x}dx\\
\text{Đặt}\ f(x) = \dfrac{1}{\sqrt{\sin x}.\cos x}>0,\ \forall x\in (0;1]\\
\qquad g(x) = \dfrac{1}{\sqrt x}>0,\ \forall x\in (0;1]\\
\text{Xét}\ \lim\limits_{x\to 0^+}\dfrac{f(x)}{g(x)}\\
\ =\lim\limits_{x\to 0^+}\dfrac{\sqrt x}{\sqrt{\sin x}.\cos x}\\
\mathop{=\kern-2pt=}\limits^{VCB} \lim\limits_{x\to 0^+}\dfrac{\sqrt x}{\sqrt x} =1\\
\Rightarrow \displaystyle\int\limits_0^1f(x)dx\ \text{và}\ \displaystyle\int\limits_0^1g(x)dx \ \text{có cùng tính chất hội tụ hoặc phân kỳ}\\
\text{Ta lại có:}\\
\displaystyle\int\limits_0^1g(x)dx= \displaystyle\int\limits_0^1\dfrac{1}{\sqrt x}dx\ \text{hội tụ}\\
\text{Do đó:}\\
\displaystyle\int\limits_0^1f(x)dx = \displaystyle\int\limits_0^1\dfrac{1}{\sqrt{\sin x}.\cos x}dx\ \text{hội tụ}\\
\text{Đặt}\ u(x) = \dfrac{1}{\sqrt{\sin x}.\cos x} >0,\ \forall x \in \left[1;\dfrac{\pi}{2}\right)\\
\qquad v(x) = \dfrac{1}{\dfrac{\pi}{2} - x}>0,\ \forall x \in \left[1;\dfrac{\pi}{2}\right)\\
\text{Xét}\ \lim\limits_{x\to \left(\tfrac{\pi}{2}\right)^-}\dfrac{u(x)}{v(x)}\\
\ = \lim\limits_{x\to \left(\tfrac{\pi}{2}\right)^-}\dfrac{\left(\dfrac{\pi}{2} - x\right)}{\sqrt{\sin x}.\cos x}\\
\ = \lim\limits_{t\to 0^-}\dfrac{-t}{\sqrt{\sin\left(t+\dfrac{\pi}{2}\right)}\cdot \cos\left(t+\dfrac{\pi}{2}\right)}\qquad \left(t = x - \dfrac{\pi}{2}\right)\\
\ = \lim\limits_{t\to 0^-}\dfrac{t}{\sqrt{\cos t}.\sin t}\\
\mathop{=\kern-2pt=}\limits^{VCB}\lim\limits_{t\to 0^-}\dfrac{t}{t} = 1\\
\Rightarrow \displaystyle\int\limits_1^{\tfrac{\pi}{2}}f(x)dx\ \text{và}\ \displaystyle\int\limits_1^{\tfrac{\pi}{2}}g(x)dx\ \text{cùng tính chất hội tụ hoặc phân kỳ}\\
\text{Ta lại có:}\\
\displaystyle\int\limits_1^{\tfrac{\pi}{2}}g(x)dx = \displaystyle\int\limits_1^{\tfrac{\pi}{2}}\dfrac{1}{\dfrac{\pi}{2} - x}dx\ \text{phân kỳ}\\
\text{Do đó:}\\
\displaystyle\int\limits_1^{\tfrac{\pi}{2}}f(x)dx = \displaystyle\int\limits_1^{\tfrac{\pi}{2}}\dfrac{1}{\sqrt{\sin x}.\cos x}dx\ \text{phân kỳ}\\
\text{Vậy tích phân đã cho phân kỳ}
\end{array}\)