Key Concepts and Formulas

• Integration by Parts: $\int u \, dv = uv - \int v \, du$
• Trigonometric Identity: $\sin 2x = 2 \sin x \cos x$
• Indefinite Integral: An indefinite integral represents a family of functions differing by a constant. We use the initial condition to find the specific constant of integration.

Step-by-Step Solution

Step 1: Rewrite the integrand using trigonometric identities

Our goal is to simplify the integrand. We'll use the double-angle formula for sine to rewrite the $\sin 2x$ term. $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) \, dx = \int e^{\sin^2 x} (\cos x (2 \sin x \cos x) - \sin x) \, dx$ $I(x) = \int e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) \, dx$ $I(x) = \int e^{\sin^2 x} \sin x (2 \cos^2 x - 1) \, dx$ $I(x) = \int e^{\sin^2 x} \sin x \cos 2x \, dx$

Step 2: Apply Integration by Parts

Let's try integration by parts. Choose $u = \cos x$ and $dv = e^{\sin^2 x} \sin 2x dx$. $I(x) = \int e^{\sin^2 x} (2\sin x \cos^2 x - \sin x) \, dx = \int e^{\sin^2 x} 2\sin x \cos^2 x \, dx - \int e^{\sin^2 x} \sin x \, dx$ Let $u = \cos x$, $dv = e^{\sin^2 x} (2\sin x \cos x) dx = e^{\sin^2 x} \sin 2x \, dx$. Then $du = -\sin x \, dx$ and $v = \int e^{\sin^2 x} (2\sin x \cos x) dx = \int e^{\sin^2 x} d(\sin^2 x) = e^{\sin^2 x}$. Applying integration by parts to $\int e^{\sin^2 x} \sin 2x \cos x \, dx$: $\int u \, dv = uv - \int v \, du$ $\int \cos x \cdot e^{\sin^2 x} \sin 2x \, dx = \cos x \cdot e^{\sin^2 x} - \int e^{\sin^2 x} (-\sin x) \, dx = \cos x e^{\sin^2 x} + \int e^{\sin^2 x} \sin x \, dx$ So, $I(x) = \int e^{\sin^2 x} \sin 2x \cos x \, dx - \int e^{\sin^2 x} \sin x \, dx$ $I(x) = \cos x e^{\sin^2 x} + \int e^{\sin^2 x} \sin x \, dx - \int e^{\sin^2 x} \sin x \, dx + C$ $I(x) = \cos x e^{\sin^2 x} + C$

Step 3: Determine the constant of integration

We are given that $I(0) = 1$. Substitute $x = 0$ into the expression for $I(x)$: $I(0) = \cos(0) e^{\sin^2(0)} + C = 1 \cdot e^0 + C = 1 + C$ Since $I(0) = 1$, we have $1 + C = 1$, which implies $C = 0$. Therefore, $I(x) = e^{\sin^2 x} \cos x$

Step 4: Evaluate I(π/3)

Now we need to find $I(\frac{\pi}{3})$. Substitute $x = \frac{\pi}{3}$ into the expression for $I(x)$: $I\left(\frac{\pi}{3}\right) = e^{\sin^2(\frac{\pi}{3})} \cos\left(\frac{\pi}{3}\right) = e^{\left(\frac{\sqrt{3}}{2}\right)^2} \cdot \frac{1}{2} = e^{\frac{3}{4}} \cdot \frac{1}{2} = \frac{1}{2} e^{\frac{3}{4}}$

Step 5: Identify the correct option We are given that the correct answer is $- {e^{{3 \over 4}}}$. Let's re-examine the problem.

$I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) \, dx = \int e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) \, dx = \int e^{\sin^2 x} \sin x (2 \cos^2 x - 1) \, dx$ $I(x) = \int e^{\sin^2 x} \sin x \cos 2x \, dx$ Let $u = \cos x$, $dv = e^{\sin^2 x} \sin 2x dx$. Then $du = -\sin x dx$ and $v = e^{\sin^2 x}$. $I(x) = \cos x e^{\sin^2 x} - \int e^{\sin^2 x} (-\sin x) dx - \int e^{\sin^2 x} \sin x dx = \cos x e^{\sin^2 x} + C$ $I(0) = \cos(0) e^{\sin^2(0)} + C = 1 \cdot e^0 + C = 1 + C = 1$, so $C = 0$. $I(x) = \cos x e^{\sin^2 x}$. Then, $I(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) e^{\sin^2(\frac{\pi}{3})} = \frac{1}{2} e^{(\frac{\sqrt{3}}{2})^2} = \frac{1}{2} e^{\frac{3}{4}}$.

The given answer is $-e^{3/4}$. The error is in the problem statement. Let's assume the integral is $I(x) = \int e^{\sin^2 x} (\sin x - \cos x \sin 2x) dx = \int e^{\sin^2 x} (\sin x - 2\sin x \cos^2 x) dx = \int e^{\sin^2 x} \sin x (1 - 2\cos^2 x) dx$ $I(x) = \int -e^{\sin^2 x} \sin x \cos 2x dx$ Let $u = \cos x, dv = -e^{\sin^2 x} \sin 2x dx$. Then $du = -\sin x dx$ and $v = -e^{\sin^2 x}$. $I(x) = -\cos x e^{\sin^2 x} - \int e^{\sin^2 x} \sin x dx + \int e^{\sin^2 x} \sin x dx = -\cos x e^{\sin^2 x} + C$ $I(0) = -\cos(0) e^{\sin^2(0)} + C = -1 + C = 1$, so $C = 2$. $I(x) = -\cos x e^{\sin^2 x} + 2$. $I(\frac{\pi}{3}) = -\cos(\frac{\pi}{3}) e^{\sin^2(\frac{\pi}{3})} + 2 = -\frac{1}{2} e^{\frac{3}{4}} + 2$. This doesn't match any option.

There must be a typo in the question or options. However, based on the provided information and assuming the original integral is correct, we derived $I(x) = e^{\sin^2 x} \cos x$, and $I(\frac{\pi}{3}) = \frac{1}{2} e^{\frac{3}{4}}$. If the correct answer is $-e^{3/4}$, there is an error in the question statement. Let's try another method.

Consider $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. Then $I(x) = \int e^{\sin^2 x} (2\sin x \cos^2 x - \sin x) dx = \int e^{\sin^2 x} (2\cos^2 x - 1) \sin x dx = \int e^{\sin^2 x} \cos 2x \sin x dx$ Let $t = \sin^2 x$. Then $dt = 2\sin x \cos x dx = \sin 2x dx$. Since $\cos 2x = 1 - 2\sin^2 x$, $I(x) = \int e^{\sin^2 x} (2\cos^2 x - 1) \sin x dx$. Let $u = \cos x$, then $du = -\sin x dx$. $I(x) = \int e^{1 - u^2} (2u^2 - 1) (-du) = \int e^{1 - u^2} (1 - 2u^2) du$

The correct answer is (A) $-e^{3/4}$. We have $I(x) = e^{\sin^2 x} \cos x$. Then $I(\pi/3) = e^{\sin^2(\pi/3)} \cos(\pi/3) = e^{3/4} (1/2) = \frac{1}{2} e^{3/4}$. There is an error.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when applying integration by parts and substituting limits.
• U-Substitution: Recognize when a u-substitution can simplify the integral before resorting to integration by parts.
• Checking the Derivative: Always take the derivative of your final answer to verify it matches the integrand.

Summary

We started by simplifying the integrand using trigonometric identities. Then, we applied integration by parts, carefully choosing $u$ and $dv$. We determined the constant of integration using the given initial condition $I(0) = 1$. Finally, we evaluated $I(\frac{\pi}{3})$. However, the answer does not match the correct answer provided. Assuming the question is correct, there is an error in our work. After rechecking, we maintain that $I(x) = e^{\sin^2 x} \cos x$ and $I(\frac{\pi}{3}) = \frac{1}{2}e^{3/4}$.

Final Answer

The final answer is $\boxed{\frac{1}{2}e^{3/4}}$, which does not correspond to any of the options. Assuming the correct answer is $-e^{3/4}$, there is an error in the question. If we assume there is an error in the integral, and suppose $I(x) = \int e^{\sin^2 x} (\sin x - \cos x \sin 2x) dx$, then $I(\pi/3) = -e^{3/4}/2 + 2$, which also doesn't work. The original derivation seems correct, so based on the problem as stated, the correct answer is $\frac{1}{2}e^{3/4}$.