If int {{e^{sec x}}left( {sec x + tan x f(x) | vedclass

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

Question

(D) Given the integral equation: $\int e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) dx = e^{\sec x} f(x) + C$. Differentiating both si

Vedclass · Accepted Answer

$\sec x + 	an x + \frac{1}{2}$

Vedclass · Answer

(D) Given the integral equation: $\int e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) dx = e^{\sec x} f(x) + C$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus: $e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) = \frac{d}{dx} (e^{\sec x} f(x))$. Applying the product rule on the right side: $e^{\sec x} (\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x) = e^{\sec x} \cdot \sec x 	an x \cdot f(x) + e^{\sec x} f'(x)$. Dividing both sides by $e^{\sec x}$: $\sec x + 	an x f(x) + \sec x 	an x + \sec^2 x = \sec x 	an x f(x) + f'(x)$. Rearranging to solve for $f'(x)$: $f'(x) = \sec x + \sec x 	an x + \sec^2 x + 	an x f(x) - \sec x 	an x f(x)$. For the equation to hold,we set the terms involving $f(x)$ to cancel out or match the structure. If we assume $f(x) = \sec x + 	an x$,then $f'(x) = \sec x 	an x + \sec^2 x$. Substituting this into the equation: $\sec x + 	an x (\sec x + 	an x) + \sec x 	an x + \sec^2 x = \sec x 	an x (\sec x + 	an x) + (\sec x 	an x + \sec^2 x)$. This simplifies to an identity. Thus,$f(x) = \sec x + 	an x + c$. Comparing with the options,$f(x) = \sec x + 	an x + \frac{1}{2}$ is a valid choice.

If $\int {{e^{\sec x}}\left( {\sec x + \tan x f(x) + (\sec x \tan x + \sec^2 x)} \right)dx = {e^{\sec x}}f(x) + C}$,then a possible choice of $f(x)$ is

Similar Questions

$\int {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \,dx = $

If $\int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}{e^{{{\cot }^{ - 1}}x}}dx = A(x) {e^{{{\cot }^{ - 1}}x}} + C}$,then $A(x)$ is equal to

Let $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$. Then find $\int e^x(f(x) + f'(x)) dx$ (where $c$ is the constant of integration).

$\int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} \right) dx = $

$\int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module