I need to find $$\int e^{-\sin^2x}{(\cos x -3 x \sin(x)+2 x \sin^3(x))}dx$$ .
I know that $$\int e^{g(x)}{(f(x)g'(x)+f'(x))}dx = e^{g(x)}f(x)$$.
But I cannot find cannot $f(x)$ in the above expression.It seems really difficult to guess what $f(x)$ could be.Any other faster methods?How do I get $f(x)$ ?
The integral is of the form \begin{equation*} \int e^{-\sin ^{2}x}\left( \cos x-3x\sin x+2x\sin ^{3}x\right) dx=\int e^{g(x)}h(x)dx. \end{equation*} The presence of the exponential function in the integrand function 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*} with $g(x)=-\sin ^{2}x.$ So we are done if we find a function $f(x)$ such that \begin{eqnarray*} h(x) &=&\left( \cos x-3x\sin x+2x\sin ^{3}x\right) \\ &=&f^{\prime }(x)+g^{\prime }(x)f(x) \\ &=&f^{\prime }(x)+\left( -2\sin x\cos x\right) f(x). \end{eqnarray*} In what follows, I will show step by step that $f(x)=x\cos x,$ and therefore \begin{equation*} \int e^{-\sin ^{2}x}\left( \cos x-3x\sin x+2x\sin ^{3}x\right) dx=x\cos xe^{-\sin ^{2}x}+C. \end{equation*} The unique thing which is given in the statement is $g(x)=-\sin ^{2}x.$ So, the first thing we start with is to look for $g^{\prime }(x)=-2\sin x\cos x$ inside what would be $f^{\prime }(x)+g^{\prime }(x)f(x),$ that is, inside \begin{equation*} \cos x-3x\sin x+2x\sin ^{3}x. \end{equation*} Since $g^{\prime }(x)=-2\sin x\cos x$ is a product of two trigonometric functions and the first two terms do not contain any product of two trig functions, then if it is possible to find $g^{\prime }(x)$ inside $h(x)$ then it is inside its third term $2x\sin ^{3}x.$ Another matter is : to get $ \cos x$ inside $\sin ^{n}x$ we need the relation $\sin ^{2}x=1-\cos ^{2}x.$ Now, it is easy to see that \begin{eqnarray*} 2x\sin ^{3}x &=&2x\sin x\sin ^{2}x \\ &=&2x\sin x(1-\cos ^{2}x) \\ &=&2x\sin x-2x\sin x\cos ^{2}x \\ &=&2x\sin x+(-2\sin x\cos x)(x\cos x) \\ &=&2x\sin x+g^{\prime }(x)(x\cos x). \end{eqnarray*} Then \begin{eqnarray*} h(x) &=&\cos x-3x\sin x+2x\sin ^{3}x \\ &=&\cos x-3x\sin x+2x\sin x+g^{\prime }(x)(x\cos x) \\ &=&\cos x-x\sin x+g^{\prime }(x)(x\cos x). \end{eqnarray*} Now we add and subtract $(x\cos x)^{\prime },$ to get \begin{equation*} h(x)=(x\cos x)^{\prime }+g^{\prime }(x)(x\cos x)+\cos x-x\sin x-(x\cos x)^{\prime } \end{equation*} But $(x\cos x)^{\prime }=\cos x-x\sin x,$ so \begin{equation*} h(x)=(x\cos x)^{\prime }+g^{\prime }(x)(x\cos x). \end{equation*} Therefore, it suffices to take $f(x)=x\cos x. \color{red} \blacksquare $
