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

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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

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$x - \sec x$

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(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:

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