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Evaluation of \(\displaystyle \int e^{x\sin x+\cos x}\cdot \left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx\)

Hi jacks.

The function \(\displaystyle{f}\) given by

\(\displaystyle{f(x)=e^{x\,\sin\,x+\cos\,x}\,\left(\dfrac{x^4\,\cos^3\,x-x\,\sin\,x+\cos\,x}{x^2\,\cos^2\,x}\right)}\)

is defined and continuous at \(\displaystyle{A=\left(\mathbb{R}-\left\{0\right\}\right)-\left\{k\,\pi+\dfrac{\pi}{2}: k\in\mathbb{Z}\right\}}\).

We integrate, for example, at \(\displaystyle{I=\left(0,\dfrac{\pi}{2}\right)}\) .

My first thought was to create the quantity \(\displaystyle{x\,\cos\,x}\) cause

\(\displaystyle{\dfrac{\mathrm{d}}{\mathrm{d}x}\,(x\,\sin\,x+\cos\,x)=x\,\cos\,x}\) .


Secondly,

\(\displaystyle{\dfrac{\mathrm{d}}{\mathrm{d}x}\,\left(\dfrac{1}{x\,\cos\,x}\right)=\dfrac{x\,\sin\,x-\cos\,x}{x^2\,\cos^2\,x}}\), so :

\(\displaystyle{\int e^{x\,\sin\,x+\cos\,x}\,\left(\dfrac{x^4\,\cos^3\,x-x\,\sin\,x+\cos\,x}{x^2\,\cos^2\,x}\right)\,\mathrm{d}x=}\)

\(\displaystyle{=\int e^{x\,\sin\,x+\cos\,x}\,\left(x^2\,\cos\,x-\left(\dfrac{1}{x\,\cos\,x}\right)'\right)\,\mathrm{d}x}\)

We have that :

\(\displaystyle{\begin{aligned} \int e^{x\,\sin\,x+\cos\,x}\,(x^2\,\cos\,x) &=\int x\,\left(e^{x\,\sin\,x+\cos\,x}\right)'\,\mathrm{d}x\\&=x\,e^{x\,\sin\,x+\cos\,x}-\int e^{x\,\sin\,x+\cos\,x}\,\mathrm{d}x\end{aligned}}\)

and

\(\displaystyle{\begin{aligned} \int e^{x\,\sin\,x+\cos\,x}\,\left(\dfrac{1}{x\,\cos\,x}\right)'\,\mathrm{d}x&=\dfrac{e^{x\,\sin\,x+\cos\,x}}{x\,\cos\,x}-\int e^{x\,\sin\,x+\cos\,x}\,\left(x\,\sin\,x+\cos\,x\right)'\,\dfrac{1}{x\,\cos\,x}\,\mathrm{d}x\\&=\dfrac{e^{x\,\sin\,x+\cos\,x}}{x\,\cos\,x}-\int e^{x\,\sin\,x+\cos\,x}\,\mathrm{d}x\end{aligned}}\)

Therefore,

\(\displaystyle{\int e^{x\,\sin\,x+\cos\,x}\,\left(\dfrac{x^4\,\cos^3\,x-x\,\sin\,x+\cos\,x}{x^2\,\cos^2\,x}\right) \,\mathrm{d}x=x\,e^{x\,\sin\,x+\cos\,x}-\dfrac{e^{x\,\sin\,x+\cos\,x}}{x\,\cos\,x}+c\,,c\in\mathbb{R}}\) .

Verification :

\(\displaystyle{\dfrac{\mathrm{d}}{\mathrm{d}x}\,\left[x\,e^{x\,\sin\,x+\cos\,x}-\dfrac{e^{x\,\sin\,x+\cos\,x}}{x\,\cos\,x}\right]=}\)

\(\displaystyle{=e^{x\,\sin\,x+\cos\,x}+x\,e^{x\,\sin\,x+\cos\,x}\,x\,\cos\,x-e^{x\,\sin\,x+\cos\,x}\,x,\cos\,x\,\dfrac{1}{x\,\cos\,x}-e^{x\,\sin\,x+\cos\,x}\,\left(\dfrac{1}{x\,\cos\,x}\right)'}\)

\(\displaystyle{=x^2\,\cos\,x\,e^{x\,\sin\,x+\cos\,x}+e^{x\,\sin\,x+\cos\,x}\,\dfrac{\cos\,x-x\,\sin\,x}{x^2\,\cos^2\,x}}\)

\(\displaystyle{=e^{x\,\sin\,x+\cos\,x}\,\left[x^2\,\cos\,x+\dfrac{\cos\,x-x\,\sin\,x}{x^2\,\cos^2\,x}\right]}\)

\(\displaystyle{=e^{x\,\sin\,x+\cos\,x}\,\left[\dfrac{x^4\,\cos^3\,x-x\,\sin\,x+\cos\,x}{x^2\,\cos^2\,x}\right]}\)