Question

The value of \[ \int_0^\frac{\pi}{2} \left( \int_0^x e^{\sin y} \sin x \, dy \right) dx \] equals ............. 
 

Hint:

For double integrals, simplify the inner integral first, and then perform the outer integration step by step.

Answer 1

Correct Answer: 1.7 - 1.8

Step 1: Interchange the order of integration

The region is \(0\le y\le x\le \frac{\pi}{2}\). Hence
\[\int_{0}^{\pi/2}\int_{0}^{x} e^{\sin y}\sin x\,dy\,dx = \int_{0}^{\pi/2}\int_{y}^{\pi/2} e^{\sin y}\sin x\,dx\,dy.\]

Step 2: Integrate with respect to \(x\)

\[\int_{y}^{\pi/2} \sin x\,dx = \big[-\cos x\big]_{y}^{\pi/2} = \cos y.\]

Thus the integral becomes
\[\int_{0}^{\pi/2} e^{\sin y}\cos y\,dy.\]

Step 3: Substitute

Let \(u=\sin y\), so \(du=\cos y\,dy\).
When \(y=0\), \(u=0\); when \(y=\frac{\pi}{2}\), \(u=1\).

\[\int_{0}^{\pi/2} e^{\sin y}\cos y \, dy = \int_{0}^{1} e^{u} \, du = e-1.\]

Final answer

\[\boxed{e-1 \approx 1.718}\]