Answer
\[ = \,1\]
Work Step by Step
\[\begin{gathered}
\int_{\frac{4}{\pi }}^\infty {\frac{1}{{{x^2}}}{{\sec }^2}\,\left( {\frac{1}{x}} \right)} dx \hfill \\
\hfill \\
set\,\,,\,\,\,u = \frac{1}{x}\,\,\,then\,\,\,du = - \frac{1}{{{x^2}}}dx\, \hfill \\
and\,\, - du = \frac{1}{{{x^2}}}dx \hfill \\
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therefore \hfill \\
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\int_{\frac{4}{\pi }}^\infty {\frac{1}{{{x^2}}}{{\sec }^2}\,\left( {\frac{1}{x}} \right)} dx = \int_{\frac{\pi }{4}}^0 {{{\sec }^2}\,\left( u \right)\,\,\,\,\left( { - du} \right)} \hfill \\
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= \int_0^{\frac{\pi }{4}} {{{\sec }^2}\,\left( u \right)du} \hfill \\
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If\,\,f\,\left( x \right)\,\,is\,\,continuous\,\,on\,\,\left( { - a,b} \right]\,\,then \hfill \\
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\int_b^b {f\,\left( x \right)} \,dx = \mathop {\lim }\limits_{c \to a} \int_c^b {f\,\left( x \right)dx} \hfill \\
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Definition\,\,of\,\,improper\,\,integral\,\,shows\,\,that. \hfill \\
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\int_{\frac{4}{\pi }}^\infty {\frac{1}{{{x^2}}}{{\sec }^2}\,\left( {\frac{1}{x}} \right)} dx\,\, = \mathop {\lim }\limits_{c \to {0^ + }} \int_0^{\frac{\pi }{4}} {{{\sec }^2}\,\left( u \right)du} \hfill \\
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= \mathop {\lim }\limits_{c \to {0^ + }} \,\,\left[ {\tan u} \right]_0^{\frac{\pi }{4}} \hfill \\
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\,\,use\,\,the\,\,fct \hfill \\
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= \mathop {\lim }\limits_{c \to {0^ + }} \,\,\left[ {\tan \frac{\pi }{4} - \tan 0} \right] \hfill \\
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evaluate\,\,\,the\,\,\lim its \hfill \\
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= 1 - \tan 0 \hfill \\
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= \,1 \hfill \\
\hfill \\
\end{gathered} \]
