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

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(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

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

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(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.

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

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