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I'm trying to integrate

$\cos\left(t\right)\,\left({\mathrm{e}}^{\cos\left(t\right)}+{\mathrm{e}}^{\sin\left(t\right)}\,\cos\left(t\right)\right)-\sin\left(t\right)\,\left({\mathrm{e}}^{\sin\left(t\right)}+{\mathrm{e}}^{\cos\left(t\right)}\,\sin\left(t\right)\right)+1$

(yeah, it's a long one)

My first thought was it's not computable, but solutions say it is

$I = t+{\mathrm{e}}^{\cos\left(t\right)}\,\sin\left(t\right)+{\mathrm{e}}^{\sin\left(t\right)}\,\cos\left(t\right)$

I don't really see how to get there since $\cos^2(t)\,e^{\cos(t)}$ can't be computed analytically.

Leon
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1 Answers1

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Let's drop the $+1$, whose effect is trivial. Now compare a rearrangement$$(\cos t-\sin^2t)e^{\cos t}+(\cos^2t-\sin t)e^{\sin t}$$with$$\frac{d}{dt}(C(t)e^{\cos t}+S(t)e^{\sin t})=(C^\prime-C\sin t)e^{\cos t}+(S^\prime+S\cos t)e^{\cos t},$$so you need to solve$$C^\prime-C\sin t=\cos t-\sin^2t,\,S^\prime+S\cos t=\cos^2t-\sin t.$$You've asked how to spot $C=\sin t,\,S=\cos t$. The phrase "by inspection" springs to mind. Since we can't solve these ODEs with a standard technique without returning to the original problem, the most natural thing to try is identifying each squared function with the term lacking a ${}^\prime$.

J.G.
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  • Well, if you know the solutions it's easy to derive. What if u're in the dark, not seeing $\frac{d}{dt}(C(t)e^{\cos t}+S(t)e^{\sin t})$ directly? It seems hard to me to spot it by inspection. – Leon May 06 '21 at 08:40
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    @Leon I hope at least the fact that we'd want to use an Ansatz of that form is intuitive, because basically the only way to get an $e^f$ factor is by differentiating $e^f$. The hard part is "inspecting" the $C,,S$ that make it work. Unfortunately, I don't think any other way of solving it exists. – J.G. May 06 '21 at 10:45