I typed
== integral (e^cosx)((cosx)^2)(sinx) from x=0 to pi
into Mathematica, but it doesn't seem to generate an answer. What did I do wrong?
I entered the query directly on the Wolfram|Alpha and there it did give an answer.
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2 Answers
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For some reason this works (in ver. 12.1) and gives the "Step-By-Step Solution" option:
WolframAlpha["--Integrate[E^Cos[x]*Cos[x]^2*Sin[x], {x, 0, Pi}]"]
but
WolframAlpha["Integrate[E^Cos[x]*Cos[x]^2*Sin[x], {x, 0, Pi}]"]
does not.
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$\begingroup$ This also works: WolframAlpha["1*Integrate[E^Cos[x]*Cos[x]^2*Sin[x], {x, 0, Pi}]"] $\endgroup$Rudy Potter– Rudy Potter2020-12-29 23:07:48 +00:00Commented Dec 29, 2020 at 23:07
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$\begingroup$ yes it works for in version 12.0 as well! $\endgroup$apri– apri2020-12-30 00:29:49 +00:00Commented Dec 30, 2020 at 0:29
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$\begingroup$ I tried == 1*integral (e^cosx) ((cosx)^2) (sinx) from x = 0 to pi, and it works now, awesome! $\endgroup$apri– apri2020-12-30 00:30:22 +00:00Commented Dec 30, 2020 at 0:30
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$\begingroup$ That's great! I'm still amazed it is ok not having a space between the trig functions and the x. $\endgroup$Rudy Potter– Rudy Potter2020-12-30 05:15:39 +00:00Commented Dec 30, 2020 at 5:15
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Alternatively, use the indefinite integral
sol = WolframAlpha[
"integral (E^(cos x))((cos x)^2)(sin x) dx", {{"IndefiniteIntegral", 2},
"Content"}, PodStates -> {"IndefiniteIntegral__Step-by-step solution"}]
From the first fundamental theorem of calculus
Subtract @@ ((-(1/2))*E^Cos[x]*
(-(4*Cos[x]) + Cos[2*x] + 5) /. {{x -> Pi}, {x -> 0}})
(* -(5/E) + E *)
Verifying,
Integrate[E^Cos[x] Cos[x]^2 Sin[x], {x, 0, Pi}]
(* -(5/E) + E *)
==. Otherwise, in Mathematica useNIntegrate[E^Cos[x] Cos[x]^2 Sin[x], {x, 0, Pi}]which is honestly not that much more complex. You can also useIntegrateinstead ofNIntegrateto receive a symbolic answer. $\endgroup$