If int e^{sin x} cdot left[ frac{x cos^3 x - | vedclass

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

Question

(A) Given the integral: $\int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} \right) dx = e^{\sin x} f(x) + c$. Simplify the integrand: $\frac

Vedclass · Accepted Answer

$x - \sec x$

Vedclass · Answer

(A) Given the integral: $\int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} ight) dx = e^{\sin x} f(x) + c$. Simplify the integrand: $\frac{x \cos^3 x - \sin x}{\cos^2 x} = x \cos x - \frac{\sin x}{\cos^2 x} = x \cos x - \sec x 	an x$. So the integral becomes: $\int e^{\sin x} (x \cos x - \sec x 	an x) dx$. We can rewrite this as: $\int [e^{\sin x} \cos x \cdot x - e^{\sin x} \sec x 	an x] dx$. Notice that $\frac{d}{dx} [e^{\sin x} (x - \sec x)] = e^{\sin x} \cos x (x - \sec x) + e^{\sin x} (1 - \sec x 	an x) = e^{\sin x} (x \cos x - \sec x \cos x + 1 - \sec x 	an x) = e^{\sin x} (x \cos x - 1 + 1 - \sec x 	an x) = e^{\sin x} (x \cos x - \sec x 	an x)$. Thus,$\int e^{\sin x} (x \cos x - \sec x 	an x) dx = e^{\sin x} (x - \sec x) + c$. Comparing this with $e^{\sin x} f(x) + c$,we get $f(x) = x - \sec x$.

If $\int e^{\sin x} \cdot \left[ \frac{x \cos^3 x - \sin x}{\cos^2 x} \right] dx = e^{\sin x} f(x) + c$,where $c$ is the constant of integration,then $f(x)$ is equal to:

Similar Questions

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

$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx}$ is equal to

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

The value of the integral $\int_{0}^{\infty} e^{-2x} (\sin 2x + \cos 2x) dx$ is:

The value of $\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module