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\( = \int ( e ^ { \sin x } \cdot \frac { \sin x } { \sec x } ) d x \)

Ask by West Weber. in Nepal
Nov 11,2024

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The integral simplifies to \( I = \frac{1}{2} \int e^{\sin x} \sin(2x) \, dx \).

Solución

To solve the integral \[ I = \int e^{\sin x} \cdot \frac{\sin x}{\sec x} \, dx, \] we first simplify the integrand. Recall that \(\sec x = \frac{1}{\cos x}\), so we can rewrite \(\frac{\sin x}{\sec x}\) as: \[ \frac{\sin x}{\sec x} = \sin x \cdot \cos x. \] Thus, the integral becomes: \[ I = \int e^{\sin x} \cdot (\sin x \cos x) \, dx. \] Next, we can use the identity \(\sin x \cos x = \frac{1}{2} \sin(2x)\) to rewrite the integral: \[ I = \int e^{\sin x} \cdot \frac{1}{2} \sin(2x) \, dx. \] This gives us: \[ I = \frac{1}{2} \int e^{\sin x} \sin(2x) \, dx. \] Now, we can use integration by parts or look for a suitable substitution. However, this integral does not have a straightforward elementary antiderivative. Instead, we can express the integral in terms of known functions or numerical methods if necessary. For practical purposes, we can leave the integral in its current form or evaluate it numerically for specific limits. Thus, the integral can be expressed as: \[ I = \frac{1}{2} \int e^{\sin x} \sin(2x) \, dx, \] and further evaluation would typically require numerical methods or special functions.

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Beyond the Answer

To tackle this integral, let’s first simplify the integrand. We can rewrite \(\sec x\) as \(\frac{1}{\cos x}\), which means \(\frac{\sin x}{\sec x} = \sin x \cdot \cos x\). Thus, the integral transforms into: \[ \int e^{\sin x} \cdot (\sin x \cdot \cos x) \, dx \] This expression can further be simplified with a substitution. You can use \(u = \sin x\), where \(du = \cos x \, dx\). So, the integral becomes: \[ \int e^u \cdot u \, du \] This is now a standard integral that can be solved using integration by parts. Integration by parts states that \(\int u \, dv = uv - \int v \, du\). Here, let \(u = u\) and \(dv = e^u du\). Thus, \(du = du\) and \(v = e^u\). Applying integration by parts gives you: \[ \int u e^u \, du = u e^u - \int e^u \, du \] This results in: \[ u e^u - e^u + C \] Finally, substitute back \(u = \sin x\) to obtain the final result: \[ \sin x \cdot e^{\sin x} - e^{\sin x} + C \] There you go! The integral simplifies beautifully and you’ve now got a nice result to show off.

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