An integration that wolfram cannot help me.

Question

$$\int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx$$

I noted the fact that $\frac{d(x\cos x)}{dx}=-x\sin x+\cos x$ but I cannot apply the substitution on it.

Answer 1

Here is the best solution I can do

Compute the following: \begin{align} & \int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx \\ & =\int e^{x\sin x+\cos x} \cdot x^2 \cos x dx+ \int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx \\ \end{align} Remark: $$\frac{d(x\sin x+\cos x)}{dx}=x \cos x$$ $$\frac{d(x\cos x)}{dx}=-x\sin x+\cos x$$ Therefore, using integration by parts, the first term is \begin{align} \int e^{x\sin x+\cos x} \cdot x^2 \cos x dx &= \int x d(e^{x\sin x+\cos x}) \\ &= x \cdot e^{x\sin x+\cos x} - \int e^{x\sin x+\cos x} dx \qquad (1) \end{align} Also, the second term is \begin{align} \int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx &= \int \frac{e^{x\sin x+\cos x}}{x^2\cos^2x}d(x \cos x) \\ &= \frac{e^{x\sin x+\cos x}}{x \cos x}-\int x \cos x d(\frac{e^{x\sin x+\cos x}}{x^2\cos^2x}) \qquad (2) \end{align} And derive: \begin{align} \int x \cos x d(\frac{e^{x\sin x+\cos x}}{x^2\cos^2x}) &= \int \frac{e^{x\sin x+\cos x} \cdot (-2 \cos x+x^2 \cos^2 x+2x \sin x)}{x^2 \cos^2 x}dx \\ &= \int e^{x\sin x+\cos x} dx - 2 \cdot \int \frac{(-x\sin x+\cos x)e^{x\sin x+\cos x}}{x^2 \cos^2 x} dx \end{align} Thus, the equation (2) becomes: \begin{align} \int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx = \frac{e^{x\sin x+\cos x}}{x \cos x} - \int e^{x\sin x+\cos x} dx \\ + 2 \cdot \int \frac{(-x\sin x+\cos x)e^{x\sin x+\cos x}}{x^2 \cos^2 x} dx \end{align} By rearranging the terms in the equation above, we have: \begin{align} -\int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx &= \frac{e^{x\sin x+\cos x}}{x \cos x} - \int e^{x\sin x+\cos x} dx \\ \int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx &= -\frac{e^{x\sin x+\cos x}}{x \cos x} + \int e^{x\sin x+\cos x} dx \qquad (3) \end{align} Combine the equation (1) and (3): \begin{align} & \int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx \\ & = x \cdot e^{x\sin x+\cos x} - \int e^{x\sin x+\cos x} dx -\frac{e^{x\sin x+\cos x}}{x \cos x} + \int e^{x\sin x+\cos x} dx \\ & = x \cdot e^{x\sin x+\cos x} -\frac{e^{x\sin x+\cos x}}{x \cos x} \end{align} as above

Answer 2

Let's write an antiderivative Ansatz $f(x)\exp (x\sin x+\cos x)$ so $$f^\prime(x) + f(x)x\cos x=x^2\cos x-\frac{1}{x}\sec x\tan x+\frac{1}{x^2}\sec x\\=x^2\cos x-\left(\frac{1}{x}\sec x\right)^\prime=1+x\cos^2 x-\left(\frac{1}{x}\sec x\right)^\prime-\frac{1}{x}\sec x\cdot x\cos x.$$But by inspection, this has solution $f(x)=x-\frac{1}{x}\sec x$.