Question

\(\int_{0}^{\frac{\pi}{2}} \sin 2x\,e^{\sin x}\,dx\) is equal to

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\)
• E. \(0\)

Hint:

For integrals involving \(\sin x\) and \(e^{\sin x}\), substitution \(t=\sin x\) simplifies the problem significantly.

Answer 1

The Correct Option is C

Step 1: Rewrite the integrand.
\[ \sin 2x=2\sin x\cos x \] So the integral becomes: \[ I=\int_{0}^{\frac{\pi}{2}}2\sin x\cos x\,e^{\sin x}\,dx \]

Step 2: Use substitution.
Let: \[ t=\sin x \] Then: \[ dt=\cos x\,dx \]

Step 3: Change the limits.
When \(x=0\), \(t=0\).
When \(x=\frac{\pi}{2}\), \(t=1\).

Step 4: Substitute into the integral.
\[ I=\int_{0}^{1}2t e^{t}\,dt \]

Step 5: Integrate by parts.
Let: \[ u=2t,\quad dv=e^t dt \] Then: \[ du=2dt,\quad v=e^t \] So, \[ I=2t e^t-\int 2e^t dt \] \[ I=2t e^t-2e^t \]

Step 6: Apply limits.
\[ I=[2t e^t-2e^t]_{0}^{1} \] At \(t=1\): \(2e-2e=0\)
At \(t=0\): \(0-2=-2\)
\[ I=0-(-2)=2 \]

Step 7: Final answer.
\[ \boxed{2} \] which matches option \((3)\).