We are tasked with evaluating the integral:

$\int e^{\frac{\sin x}{3}} \sin 2x \, dx$

and using the substitution $t = e^{\frac{\sin x}{3}}$. Let’s proceed step by step.

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Step 1: Apply the substitution

Let: $t = e^{\frac{\sin x}{3}}$

Taking the derivative with respect to $x$, we get: $\frac{dt}{dx} = e^{\frac{\sin x}{3}} \cdot \frac{1}{3} \cos x \quad \implies \quad dt = \frac{1}{3} e^{\frac{\sin x}{3}} \cos x \, dx$

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Step 2: Rewrite $\sin 2x$

Recall the identity: $\sin 2x = 2 \sin x \cos x$

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Step 3: Substitute into the integral

The integral becomes: $\int e^{\frac{\sin x}{3}} \sin 2x \, dx = \int t \cdot 2 \sin x \cos x \, dx$

From $dt = \frac{1}{3} t \cos x \, dx$, we isolate $\sin x \cos x \, dx$ as follows: $\sin x \cos x \, dx = \frac{3}{2} \, dt$

Substitute into the integral: $\int e^{\frac{\sin x}{3}} \sin 2x \, dx = \int t \cdot \frac{3}{2} \, dt$

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Step 4: Simplify and integrate

Factor out constants: $\int t \cdot \frac{3}{2} \, dt = \frac{3}{2} \int t \, dt$

The integral of $t$ with respect to $t$ is: $\frac{3}{2} \cdot \frac{t^2}{2} = \frac{3}{4} t^2 + C$

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Step 5: Substitute back $t$

Since $t = e^{\frac{\sin x}{3}}$, we have: $\int e^{\frac{\sin x}{3}} \sin 2x \, dx = \frac{3}{4} \left(e^{\frac{\sin x}{3}}\right)^2 + C$

Simplify: $\int e^{\frac{\sin x}{3}} \sin 2x \, dx = \frac{3}{4} e^{\frac{2 \sin x}{3}} + C$

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Final Answer:

$\boxed{\int e^{\frac{\sin x}{3}} \sin 2x \, dx = \frac{3}{4} e^{\frac{2 \sin x}{3}} + C}$

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Do you want further clarification on any step or a related explanation? 😊

5 Related Questions:

1. How is the substitution $t = e^{\frac{\sin x}{3}}$ determined?
2. Why do we rewrite $\sin 2x$ using trigonometric identities?
3. Can substitution methods always simplify exponential integrals?
4. What is the general approach to solving integrals involving $e^{\sin x}$?
5. How does the constant factor affect the integral solution?

Tip:

Always verify your substitution by differentiating back to ensure the limits or terms align perfectly!