$$\frac{\mathrm d}{\mathrm dx}\int_1^x \mathrm e^{\cos t}\,\mathrm dt$$ I get the solution $\mathrm e^{\cos x}$ however it is not the exact solution. The correct solution $\mathrm e^{\cos t}$.
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The integral has no closed form solution. Check out https://math.stackexchange.com/questions/2468863/what-is-the-integral-of-e-cos-x – Tony Hellmuth Jun 01 '18 at 05:08
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@TonyHellmuth: Note that since we're differentiating the function, we don't really care about a closed form solution for the integral. – Richard Ambler Jun 01 '18 at 05:27
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I think that your solution is correct; $e^{\cos t}$ given as a solution must be a mistake in the book. The variable $t$ is a dummy variable that only has meaning within the integral.
Richard Ambler
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d/d∫e^cosd =d/dx[e^cos/sint] for 1,x
d/d∫e^cosd = d/dx[e^cosx/sinx-e^cos1/sin1] = -sinxe^cosx/sinx= - e^cosx
so it is probably a mistake from you book first you have to integrate and then differentiate in order to get the result
