int e^{sin x} frac{(x cos^3 x - sin x)}{cos^2 | vedclass

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Question

(A) Let $I = \int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} \right) dx$ $= \int e^{\sin x} (x \cos x - \frac{\sin x}{\cos^2 x}) dx$ $= \i

Vedclass · Accepted Answer

$e^{\sin x}(x - \sec x) + C$

Vedclass · Answer

(A) Let $I = \int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} ight) dx$ $= \int e^{\sin x} (x \cos x - \frac{\sin x}{\cos^2 x}) dx$ $= \int e^{\sin x} (x \cos x - 	an x \sec x) dx$ $= \int x (e^{\sin x} \cos x) dx - \int e^{\sin x} \sec x 	an x dx$ Using integration by parts for the first integral,let $u = x$ and $dv = e^{\sin x} \cos x dx$. Then $du = dx$ and $v = e^{\sin x}$. $= x e^{\sin x} - \int e^{\sin x} dx - \int e^{\sin x} \sec x 	an x dx$ This approach is complex. Let's rewrite the integrand: $I = \int e^{\sin x} (x \cos x - \sec x 	an x) dx$ $= \int x e^{\sin x} \cos x dx - \int e^{\sin x} \sec x 	an x dx$ $= x e^{\sin x} - \int e^{\sin x} dx - \int e^{\sin x} \sec x 	an x dx$ Actually,consider $f(x) = x e^{\sin x}$. Then $f'(x) = e^{\sin x} + x e^{\sin x} \cos x$. Consider $g(x) = -e^{\sin x} \sec x$. Then $g'(x) = -[e^{\sin x} \cos x \sec x + e^{\sin x} \sec x 	an x] = -[e^{\sin x} + e^{\sin x} \sec x 	an x]$. Thus,$\frac{d}{dx} [e^{\sin x} (x - \sec x)] = e^{\sin x} \cos x (x - \sec x) + e^{\sin x} (1 - \sec x 	an x) = x e^{\sin x} \cos x - e^{\sin x} + e^{\sin x} - e^{\sin x} \sec x 	an x = e^{\sin x} (x \cos x - \sec x 	an x)$. Therefore,the integral is $e^{\sin x}(x - \sec x) + C$.

$\int e^{\sin x} \frac{(x \cos^3 x - \sin x)}{\cos^2 x} dx =$

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