int {{e^{sin x}}left( {sin x + {{sec }^2}x} r | vedclass

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(A) We are given the integral $I = \int e^{\sin x} (\sin x + \sec^2 x) \, dx$. This integral does not fit the standard form $\int e^x (f(x) + f'(x)) \

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${e^{\sin x}}	an x + C$

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(A) We are given the integral $I = \int e^{\sin x} (\sin x + \sec^2 x) \, dx$. This integral does not fit the standard form $\int e^x (f(x) + f'(x)) \, dx$ directly because the exponent is $\sin x$ and not $x$. Let $u = \sin x$,then $du = \cos x \, dx$,which implies $dx = \frac{du}{\cos x}$. The integral becomes $I = \int e^u (u + \sec^2 x) \frac{du}{\cos x}$. Since $\cos x = \sqrt{1 - u^2}$,we have $I = \int e^u \left( \frac{u}{\sqrt{1 - u^2}} + \frac{1}{(1 - u^2)^{3/2}} ight) \, du$. This does not simplify to a standard elementary form. Checking the options provided,none of the functions $f(x)$ such that $\frac{d}{dx} [e^{\sin x} f(x)] = e^{\sin x} (\sin x + \sec^2 x)$ exist. Specifically,$\frac{d}{dx} [e^{\sin x} 	an x] = e^{\sin x} \cos x 	an x + e^{\sin x} \sec^2 x = e^{\sin x} (\sin x + \sec^2 x)$. Wait,$\cos x 	an x = \sin x$. Therefore,$\frac{d}{dx} [e^{\sin x} 	an x] = e^{\sin x} \sin x + e^{\sin x} \sec^2 x = e^{\sin x} (\sin x + \sec^2 x)$. Thus,the integral is $e^{\sin x} 	an x + C$.

$\int {{e^{\sin x}}\left( {\sin x + {{\sec }^2}x} \right)} \,dx$ is equal to

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