Integral problem

Question

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 $\int e^t(f(t)+f'(t)) \, dt$

Answer 1

\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...)

Answer 2

You have the answer (without solution) as an answer and a very god hint. Here goes a way of thinking:

If we should have any chance to get this one, I think the primitive has to be in the form $$ e^{x\sin x+\cos x}f(x) $$ for some function $f$. Moreover, $$ De^{x\sin x+\cos x}f(x)=e^{x\sin x+\cos x}(f'(x)+x\cos x f(x)), $$ so our $f$ must satisfy $$ f'(x)+x\cos x f(x)=\frac{x^4\cos^3 x-x \sin x+\cos x}{x^2\cos^2 x}. $$ We observe that he first term in the right-hand side reads (upon division) $x^2\cos x$. By comparing, this suggests that our function $f$ could be written $$ f(x)=x+g(x) $$ for some function $g$. But then $g$ should satisfy $$ 1+g'(x)+x\cos x g(x)=\frac{-x \sin x+\cos x}{x^2\cos^2 x}, $$ or, moving the $1$ to the right-hand side, $$ g'(x)+x\cos x g(x)=\frac{-x \sin x+\cos x}{x^2\cos^2 x}-1. $$ Next, the $x^2\cos^2x$ in the denominator suggests that the function $g$ can be written $$ g(x)=\frac{h(x)}{x\cos x}, $$ for some function $h$. Differentiating gives $$ g'(x)+x\cos x g(x)=\frac{h'(x)x\cos x-h(x)(\cos x-x\sin x)}{x^2\cos^2x}+h(x). $$ Very lucky shot! With $h(x)=-1$, we are all set. Thus, a primitive is $$ e^{x\sin x+\cos x}\Bigl(x-\frac{1}{x\cos x}\Bigr) $$

Answer 3

The integral is of the form \begin{equation*} \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=\int e^{g(x)}h(x)dx. \end{equation*} This form recalls the well-known formula \begin{equation*} \int e^{g(x)}\left( f^{\prime }(x)+g^{\prime }(x)f(x)\right) dx=f(x)e^{g(x)}+C. \end{equation*}

Its proof maybe found at

Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$

So we are done if we find a function $f(x)$ such that \begin{equation*} h(x)=\frac{x^{4}\cos ^{3}x-x\sin x+\cos x}{x^{2}\cos ^{2}x}=f^{\prime }(x)+g^{\prime }(x)f(x)=f^{\prime }(x)+(x\cos x)f(x) \end{equation*} In what follows, I will show that $f(x)=x-\frac{1}{x\cos x}$ and therefore \begin{eqnarray*} \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 &=&\left( f(x)\right) e^{g(x)}+C \\ &=&\left( x-\frac{1}{x\cos x}\right) e^{x\sin x+\cos x}+C. \end{eqnarray*} Remark. If $f(x)=f_{1}(x)+f_{2}(x)$ then \begin{eqnarray*} f^{\prime }(x)+g^{\prime }(x)f(x) &=&(f_{1}+f_{2})^{\prime }+g^{\prime }(x)(f_{1}+f_{2}) \\ &=&\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +\left( f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) \end{eqnarray*} conversely, if \begin{equation*} h(x)=\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +\left( f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) \end{equation*} then \begin{equation*} h(x)=f^{\prime }(x)+g^{\prime }(x)f(x),\ \ with\ \ f(x)=f_{1}(x)+f_{2}(x). \end{equation*} This remark suggests to look for $f$ by pieces! that is, if we can write \begin{equation*} h(x)=\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +something \end{equation*} we reduce our task to look for $f_{2}$ such that \begin{equation*} something=\left( f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) \end{equation*} and therefore $h(x)$ would be $(f_{1}+f_{2})^{\prime }+g^{\prime }(x)(f_{1}+f_{2}),$ that is we can take $f=f_{1}+f_{2}.$

The procedure can be described as follows. First look for $g^{\prime }(x)$ inside $h(x).$ If some expression like $g^{\prime }(x)f_{1}(x)$ is found \begin{equation*} h(x)=g^{\prime }(x)f_{1}(x)+something \end{equation*} then add and subtract $f_{1}^{\prime }(x)$ and write \begin{equation*} h(x)=\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +something-f_{1}^{\prime }(x). \end{equation*} Next take $h_{1}(x)=something-f_{1}^{\prime }(x)$ and try to find $g^{\prime }(x)f_{2}(x)$ inside $h_{1}(x)$. If found, add and subtract $f_{2}^{\prime }(x).$ And so on. Since the integral to be evaluated is a reasonable integral which come from an exercise textbook, then at some moment this procedure should stop, for example when $f_{n}(x)$ is obtained, and therefore \begin{equation*} h(x)=\sum_{k=1}^{n}\left( f_{k}^{\prime }(x)+g^{\prime }(x)f_{k}(x)\right) \end{equation*} and \begin{equation*} f(x)=\sum_{k=1}^{n}f_{k}(x). \end{equation*}

According to my own experience, $n=1$ or $2.$ I never get $n=3.$

Let's go.!

The unique thing which is given in the statement is $g(x)=x\sin x+\cos x.$ So, the first thing we start with is to look for $g^{\prime }(x)=x\cos x$ inside what would be $h(x)=f^{\prime }(x)+g^{\prime }(x)f(x),$ that is, inside \begin{equation*} \frac{x^{4}\cos ^{3}x-x\sin x+\cos x}{x^{2}\cos ^{2}x}. \end{equation*} If we separate the fraction into 3 fractions we get \begin{equation*} x^{2}\cos x-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}. \end{equation*} It is easy to see $g^{\prime }(x)=x\cos x$ in the first term \begin{equation*} xg^{\prime }(x)-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}. \end{equation*} This suggests to take $f_{1}(x)=x.$ So, we have to add and subtract $ f_{1}^{\prime }(x)$ to get the package $\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) ,$ \begin{equation*} \left( x^{\prime }+g^{\prime }(x)x\right) -\frac{\sin x}{x\cos ^{2}x}+\frac{1 }{x^{2}\cos x}-1. \end{equation*} Now we take \begin{equation*} h_{2}(x)=-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}-1 \end{equation*} and try to find inside it $g^{\prime }(x)=x\cos x.$ This one is not present in the first two fractions, but can be in the third as follows \begin{eqnarray*} h_{2}(x) &=&-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}+(x\cos x)\left( \frac{-1}{x\cos x}\right) \\ &=&-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}+g^{\prime }(x)\left( \frac{-1}{x\cos x}\right) . \end{eqnarray*} So we take $f_{2}(x)=\left( \frac{-1}{x\cos x}\right) ,$ and we have to add and subtract $f_{2}^{\prime }(x)$ \begin{eqnarray*} h_{2}(x) &=&\left( \left( \frac{-1}{x\cos x}\right) ^{\prime }+g^{\prime }(x)\left( \frac{-1}{x\cos x}\right) \right) -\frac{\sin x}{x\cos ^{2}x}+ \frac{1}{x^{2}\cos x}-\left( \frac{-1}{x\cos x}\right) ^{\prime } \\ &=&\left( f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) -\frac{\sin x}{ x\cos ^{2}x}+\frac{1}{x^{2}\cos x}-\left( \frac{-1}{x\cos x}\right) ^{\prime } \end{eqnarray*} However, easy computation shows that \begin{equation*} -\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}-\left( \frac{-1}{x\cos x} \right) ^{\prime }=0. \end{equation*} (This happens because the exercise is from a textbook) Therefore \begin{equation*} h(x)=f_{1}(x)+f_{2}(x)=x-\frac{1}{x\cos x} \end{equation*} and \begin{equation*} \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=\left( x-\frac{1}{x\cos x}\right) e^{x\sin x+\cos x}+C. \ \ \ \color{red} \blacksquare \end{equation*}