<s>[INST] Integral problem Find $$ \int e^{x \sin x+\cos x} \left(\frac{x^4\cos^3 x-x \sin x+\cos x}{x^2\cos^2 x}\right) \, dx$$
My attempt:I tried putting $x \sin x+\cos x=t$ and cannot express it in the form of $\int e^t(f(t)+f'(t)) \, dt$
 [/INST] \begin{align}
& \int e^{x\sin x+\cos x}(\frac{x^4\cos^3x-x\sin^2x+\cos x}{x^2\cos^2x})dx\\
& \hspace{5mm} =\int e^{x\sin x+\cos x}(x^2\cos x-\frac{x\sin^2x-\cos x}{x^2\cos^2x})dx\\
& \hspace{5mm} =\int e^{x\sin x+\cos x}(x^2\cos x-\frac{x\tan^2x-\sec x}{x^2})dx
\end{align}
Realize that $\frac{x\tan^2x-\sec x}{x^2}=\frac{d}{dx}\frac{\sec x}{x}$ and that you can make $\frac{\sec x}{x}$ appear elsewhere by factoring $x^2\cos x-1$ into $(x-\frac{\sec x}{x})(x\cos x)$. So the above is equal to:
\begin{align}
& \int e^{x\sin x+\cos x} \left((x-\frac{\sec x}{x})(x\cos x)+1-\frac{x\tan^2x-\sec x}{x^2}\right) \, dx\\
&=\int \left[e^{x\sin x+\cos x}\left(x-\frac{\sec x}{x}\right)(x\cos x)+e^{x\sin x+\cos x}\left(1-\frac{x\tan^2x-\sec x}{x^2}\right)\right]dx
\end{align}
Now realize that $e^{x\sin x+\cos x}(x\cos x)=\frac{d}{dx}e^{x\sin x+\cos x}$. The above is equal to:
$$\int \left[\left(x-\frac{\sec x}{x}\right)\frac{d}{dx}(e^{x\sin x+\cos x})+e^{x\sin x+\cos x}\frac{d}{dx}\left(x-\frac{\sec x}{x}\right)\right]\,dx
$$
Now, this looks exactly looks like the product rule with $u=x-\frac{\sec x}{x}$ and $v=e^{x\sin x+\cos x}$. So the integral is equal to $$(x-\frac{\sec x}{x})e^{x\sin x+\cos x}+C$$
(To be honest, I did use WolframAlpha to evaluate the integral and work backward to take the derivative by hand, and then reverse each step, but I don't see any other way of evaluating such a difficult integral by hand...)
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