Evaluating integral $\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $

Question

$$\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $$ The usual form $\int e^x(f(x)+f'(x))dx $ does not apply here. What substitution should I make ?

Answer 1

This is a tricky one! We spot a lot of things here that look like derivatives, so we should try to reframe our expression to reflect that, and make it useful for us. Split it up as follows:

$$[e^{\cos x}\sin x]x+e^{\cos x}[\frac{\cos x}{\sin^2 x}]$$

Let's integrate by parts here, setting $u'=e^{\cos x}\sin x, \frac{\cos x}{\sin^2 x}$ and $v=x, e^{\cos x}$ respectively. Noting that $\int u'v\ dx = uv - \int uv'\ dx$, we see:

$$\int [e^{\cos x}\sin x]xdx=(-e^{\cos x})x-\int[-e^{\cos x}]dx$$

$$\int e^{\cos x}[\frac{\cos x}{\sin^2 x}]dx=(e^{\cos x})(\frac{-1}{\sin x})-\int[-e^{\cos x}\sin x](\frac{-1}{\sin x})dx$$

We then notice that the extra integrals we pick up cancel each other out perfectly, leaving us with:

$$-(x+\frac{1}{\sin x})e^{\cos x}+C$$

Answer 2

The integral is of the form \begin{equation*} \int \left( x\sin x+\frac{\cos x}{\sin ^{2}x}\right) e^{\cos x}dx=\int h(x)e^{g(x)}dx. \end{equation*} This form recalls the well-known formula \begin{equation*} \int \left( f^{\prime }(x)+g^{\prime }(x)f(x)\right) e^{g(x)}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)=f^{\prime }(x)+g^{\prime }(x)f(x). \end{equation*} In what follows, I will show that $f(x)=-x-\csc x,$ and therefore \begin{equation*} \int \left( x\sin x+\frac{\cos x}{\sin ^{2}x}\right) e^{\cos x}dx=\left( -x-\csc x\right) e^{\cos x}+C. \end{equation*} $\color{red}{\bf Problem:}$ We want to write $x\sin x+\frac{\cos x}{\sin ^{2}x}$ as $ f^{\prime }(x)+g^{\prime }(x)f(x)$ where $g(x)=\cos x,$ $g^{\prime }(x)=-\sin x$ and $f(x)$ is to be determined.

First, it is easy to see that \begin{equation*} x\sin x+\frac{\cos x}{\sin ^{2}x}=\frac{\cos x}{\sin ^{2}x}+(-\sin x)(-x)= \frac{\cos x}{\sin ^{2}x}+g^{\prime }(x)(-x). \end{equation*} If we put $f_{1}(x)=(-x),$ then \begin{equation*} f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)=\left( -1\right) +(-\sin x)\left( -x\right) =-1+x\sin x \end{equation*} This suggests to add and to subtract the term $-1$ as follows \begin{equation*} x\sin x+\frac{\cos x}{\sin ^{2}x}=x\sin x+\frac{\cos x}{\sin ^{2}x}% -1+1=\left( -1+x\sin x\right) +\left( 1+\frac{\cos x}{\sin ^{2}x}\right) , \end{equation*} Now let us find $f_{2}(x)$ such that : \begin{equation*} \left( 1+\frac{\cos x}{\sin ^{2}x}\right) =f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)=f_{2}^{\prime }(x)-(\sin x)f_{2}(x). \end{equation*} It is easy to see that \begin{equation*} \frac{\cos x}{\sin ^{2}x}+1=\frac{\cos x}{\sin ^{2}x}+(-\sin x)\left( \frac{ -1}{\sin x}\right) =\frac{\cos x}{\sin ^{2}x}+g^{\prime }(x)\left( \frac{-1}{ \sin x}\right) \end{equation*} Then if we put $f_{2}(x)=\left( \frac{-1}{\sin x}\right) $ it follows that \begin{equation*} f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)=\left( \frac{-1}{\sin x}\right) ^{\prime }-(\sin x)\left( \frac{-1}{\sin x}\right) =\frac{\cos x}{\sin ^{2}x} +1. \end{equation*} It follows that \begin{eqnarray*} \left( x\sin x+\frac{\cos x}{\sin ^{2}x}\right) &=&\left( -1+x\sin x\right) +\left( \frac{\cos x}{\sin ^{2}x}+1\right) \\ && \\ &=&\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +\left( f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) \\ && \\ &=&\left( (f_{1}(x)+f_{2}(x))^{\prime }+g^{\prime }(x)(f_{1}(x)+f_{2}(x))\right) \end{eqnarray*} then, it suffices to take

\begin{equation*} f(x)=f_{1}(x)+f_{2}(x)=-x-\frac{1}{\sin x}=-x-\csc x.\ \ \ \color{red} \blacksquare \end{equation*}