int e^{sin x} cdot left(frac{sin x+1}{sec x} | vedclass

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Question

(A) Let $I = \int e^{\sin x} \cdot \left(\frac{\sin x+1}{\sec x}\right) d x$.Since $\frac{1}{\sec x} = \cos x$,we can rewrite the integral as:$I = \in

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int e^{\sin x} \cdot \left(\frac{\sin x+1}{\sec x}\right) d x$.Since $\frac{1}{\sec x} = \cos x$,we can rewrite the integral as:$I = \int e^{\sin x} (\sin x + 1) \cos x \, dx$.Let $u = \sin x$,then $du = \cos x \, dx$.Substituting these into the integral:$I = \int e^u (u + 1) \, du$.$I = \int (u e^u + e^u) \, du$.Using the integration by parts formula $\int (f(u) + f'(u)) e^u \, du = e^u f(u) + C$,where $f(u) = u$ and $f'(u) = 1$:$I = e^u \cdot u + C$.Substituting $u = \sin x$ back:$I = \sin x \cdot e^{\sin x} + C$.

$ \int e^{\sin x} \cdot \left(\frac{\sin x+1}{\sec x}\right) d x $ is equal to

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