\(\int_{0}^{\pi}\displaystyle e^{\sin x}\cos x\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Let \(t = \sin x\), so \(dt = \cos x\,dx\).
When \(x = 0\), \(t = \sin 0 = 0\); when \(x = \pi\), \(t = \sin\pi = 0\).
\( \displaystyle \int_{0}^{\pi} e^{\sin x}\cos x\,dx = \int_{0}^{0} e^{t}\,dt \)
\( = 0\)
Both limits become 0, and an integral with equal limits is zero.
Why the other options are incorrect
Option (1) \(e-1\) — Incorrect. This is the value over \([0, \pi/2]\) only; over \([\pi/2, \pi]\) the integral is \(1-e\), which cancels it.
Option (2) \(2(e-1)\) — Incorrect. It adds the two halves as if both were positive, but \(\cos x \lt 0\) on \((\pi/2, \pi)\).
Option (4) \(e+1\) — Incorrect. The limits of t are both 0, so the value is \(e^{0}-e^{0} = 0\).