int cos ^{-1}left(sqrt{frac{x}{a+x}}right) d | vedclass

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

Question

(D) Given that $\int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = f(x) + C$. By the Fundamental Theorem of Calculus,the derivative of the integral is the int

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) Given that $\int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = f(x) + C$. By the Fundamental Theorem of Calculus,the derivative of the integral is the integrand: $f^{\prime}(x) = \frac{d}{dx} \int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = \cos ^{-1} \sqrt{\frac{x}{a+x}}$. Now,we need to find $f^{\prime}(a)$. Substitute $x = a$ into the expression for $f^{\prime}(x)$: $f^{\prime}(a) = \cos ^{-1} \sqrt{\frac{a}{a+a}} = \cos ^{-1} \sqrt{\frac{a}{2a}}$. Simplifying the fraction inside the square root: $f^{\prime}(a) = \cos ^{-1} \sqrt{\frac{1}{2}} = \cos ^{-1} \left(\frac{1}{\sqrt{2}}\right)$. Since $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$,we have: $f^{\prime}(a) = \frac{\pi}{4}$.

$\int \cos ^{-1}\left(\sqrt{\frac{x}{a+x}}\right) d x=f(x)+C \Rightarrow f^{\prime}(a)=$

Similar Questions

$\int \frac{5(x^6+1)}{x^2+1} \, dx =$ (Where $C$ is a constant of integration.)

$\int \left(\sum_{r=0}^{\infty} \frac{x^r 3^r}{r!}\right) dx =$

Evaluate the integral: $\int a^{3x + 3} dx$.

$\int \frac{a^{\sqrt{x}}}{\sqrt{x}} dx = $

$\int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$ is equal to (where $c$ is a constant of integration).

Vedclass Test Series

Exam Paper Generator

Online Exam Module