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\(\int_0^{\pi / 2} e^{\sin x} \cdot \cos x d x=\)
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Verified Answer
The correct answer is:
\(e-1\)
\(I=\int_0^{\pi / 2} e^{\sin x} \cos x d x\)
Put \(\sin x=t\), then at \(x=0, t=0\) and at \(x=\frac{\pi}{2}, t=1\) and \(\cos x d x=d t\)
So, \(I=\int_0^1 e^t d t=\left[e^t\right]_0^1=e^1-1=e-1\)
Put \(\sin x=t\), then at \(x=0, t=0\) and at \(x=\frac{\pi}{2}, t=1\) and \(\cos x d x=d t\)
So, \(I=\int_0^1 e^t d t=\left[e^t\right]_0^1=e^1-1=e-1\)
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