int e^x left( frac{sec^2 x + tan x - cot x}{s | vedclass

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

Question

(D) Let $I = \int e^x \left( \frac{\sec^2 x + 	an x - \cot x}{\sin x} ight) dx$.We can rewrite the integrand as:$I = \int e^x \left( \frac{\sec^2 x

Vedclass · Accepted Answer

$e^x(\operatorname{cosec} x + \sec x) + c$

Vedclass · Answer

(D) Let $I = \int e^x \left( \frac{\sec^2 x + 	an x - \cot x}{\sin x} ight) dx$.We can rewrite the integrand as:$I = \int e^x \left( \frac{\sec^2 x}{\sin x} + \frac{	an x}{\sin x} - \frac{\cot x}{\sin x} ight) dx$$I = \int e^x \left( \sec^2 x \operatorname{cosec} x + \sec x - \cot x \operatorname{cosec} x ight) dx$Using the identity $\sec^2 x = 1 + 	an^2 x$, we have:$I = \int e^x \left( \operatorname{cosec} x(1 + 	an^2 x) + \sec x - \cot x \operatorname{cosec} x ight) dx$$I = \int e^x \left( \operatorname{cosec} x + \operatorname{cosec} x 	an^2 x + \sec x - \cot x \operatorname{cosec} x ight) dx$Since $\operatorname{cosec} x 	an^2 x = \frac{1}{\sin x} \cdot \frac{\sin^2 x}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x 	an x$, the expression becomes:$I = \int e^x \left( (\operatorname{cosec} x - \cot x \operatorname{cosec} x) + (\sec x + \sec x 	an x) ight) dx$This is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$.Here, $f(x) = \operatorname{cosec} x + \sec x$, and $f'(x) = -\operatorname{cosec} x \cot x + \sec x 	an x$.Thus, $I = e^x(\operatorname{cosec} x + \sec x) + c$.Therefore, option $(D)$ is correct.

$\int e^x \left( \frac{\sec^2 x + \tan x - \cot x}{\sin x} \right) dx =$

Similar Questions

The value of $\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$ is equal to

The solution of $\frac{dy}{dx} = e^x(\sin x + \cos x)$ is

The value of $\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x+\cot x) d x$ is

$\int e^{-x}(x^3-2x^2+3x-4) dx=$

$\int {\frac{{x - 1}}{{{{(x + 1)}^3}}}{e^x}\,dx = } $

Vedclass Test Series

Exam Paper Generator

Online Exam Module