Let int {{sec }^{ - 1}}left[ { - {sin }^2x} r | vedclass

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Question

(A) Given the integral $I = \int {{\sec }^{ - 1}}\left[ { - {\sin }^2x} \right]dx$. Since $0 \le {\sin }^2x \le 1$,we have $-1 \le -{\sin }^2x \le 0$.

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$2$

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(A) Given the integral $I = \int {{\sec }^{ - 1}}\left[ { - {\sin }^2x} ight]dx$. Since $0 \le {\sin }^2x \le 1$,we have $-1 \le -{\sin }^2x \le 0$. For $x 
eq 0$,$-{\sin }^2x$ is in the interval $[-1, 0)$. The greatest integer function $[ - {\sin }^2x ]$ is $-1$ for all $x 
eq n\pi$. Thus,the integral becomes $\int {{\sec }^{ - 1}}( - 1)dx = \int \pi dx = \pi x + C$. Given $f(0) = 0$,we have $f(x) = \pi x$. Now,we need to find the value of ${\left( {f\left( {\frac{8}{{\pi x}}} ight)} ight)''}$ at $x = 2$. $f\left( {\frac{8}{{\pi x}}} ight) = \pi \left( {\frac{8}{{\pi x}}} ight) = \frac{8}{x} = 8x^{-1}$. First derivative: $\frac{d}{dx}(8x^{-1}) = -8x^{-2}$. Second derivative: $\frac{d^2}{dx^2}(8x^{-1}) = 16x^{-3} = \frac{16}{x^3}$. At $x = 2$,the value is $\frac{16}{2^3} = \frac{16}{8} = 2$.

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