Question

If I(x) = \(\int e^{sin^2x}\)\((cos \;x \;sin \;2x - sin \;x)dx\) and I (0) = 1, then \(I(π/3)\) is equal to

• A. \(-e^{\frac{3}{4}}\)
• B. \(e^{\frac{3}{4}}\)
• C. \(-\frac{1}{2}e^{\frac{3}{4}}\)
• D. \(\frac{1}{2}e^{\frac{3}{4}}\)

Answer 1

The Correct Option is D

Given:

\( I = \int e^{\sin^2x} \sin^2x \cos x \, dx - \int e^{\sin^2x} \sin x \, dx \).

• Step 1: Simplify the expression:

  \( I = \cos x \int e^{\sin^2x} \sin^2x \, dx - (-\sin x) \int e^{\sin^2x} \sin^2x \, dx - \int e^{\sin^2x} \sin x \, dx \).

  Substitute \( \sin^2x = t \), so \( 2\sin x \cos x \, dx = dt \):

  \( I = \int e^t \, dt + \int e^t \, dt - \int e^{\sin^2x} \sin x \, dx \).

• Step 2: Solve the integration:

  \( I = e^{\sin^2x} \cos x + e^{\sin^2x} \sin x \, dx - e^{\sin^2x} \sin x \, dx \).

  Simplify:

  \( I = e^{\sin^2x} \cos x + C \).

• Step 3: Apply the initial condition \( I(0) = 1 \):

  \( I(0) = e^{\sin^2(0)} \cos(0) + C \).

  Simplify:

  \( 1 = 1 + C \implies C = 0 \).

  Thus:

  \( I = e^{\sin^2x} \cos x \).

• Step 4: Find \( I\left(\frac{\pi}{3}\right) \):

  \( I\left(\frac{\pi}{3}\right) = e^{\sin^2\left(\frac{\pi}{3}\right)} \cos\left(\frac{\pi}{3}\right) \).

  Substitute values:

  \( I\left(\frac{\pi}{3}\right) = e^{\left(\frac{\sqrt{3}}{2}\right)^2} \cdot \frac{1}{2} \).

  Simplify:

  \( I\left(\frac{\pi}{3}\right) = e^{\frac{3}{4}} \cdot \frac{1}{2} = \frac{e^{\frac{3}{4}}}{2} \).

Final Answer: \( I\left(\frac{\pi}{3}\right) = \frac{e^{\frac{3}{4}}}{2} \).