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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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 $\frac{16}{2^3} = \frac{16}{8} = 2$.

Similar Questions

If $\int \frac{\sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} dx = -\frac{f(x)}{2} + c$,then the domain of $f(x)$ is

$\int \tan ^{-1}\left(1-x+x^2\right) d x+\int \tan ^{-1}(x) d x+\int \tan ^{-1}(1-x) d x=$

If $\int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^6}} \right)}^{2/3}}}} = xf\left( x \right){{\left( {1 + {x^6}} \right)}^{\frac{1}{3}}} + C} $ where $C$ is a constant of integration,then the function $f(x)$ is equal to

For $x \geq 0$, $\int \sqrt{x^2+2x} \, dx$ is equal to

If $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$, then $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module