Tripos Integral

Show that $$\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{e^{2mx}(\cos x-m\sin x)^2}dx=\frac{1}{2m(1+m^2)}\left[\frac{1+m}{1-m}e^{-m\frac{\pi}{2}}-1\right]$$ This problem is from Edwards, Treatise on Integral Calculus II, pg.187.

It is a definite integral and is not likely to find the indefinite integral, at this point in the book Edwards has introduces the tricks of substituting $\frac{\pi}{4}-x$ for $x$, and expansion in series (and of course parts). Unfortunately, I haven't been able to make these tricks work out. My method has been to expand $\frac{\tan^2 x}{(1-m\tan x)^2}$ as a series in $m\tan x$, and integrate termwise but so far without success. Does anybody have any good ideas?

Solution 1:

We'll first remove the trigonometric functions since it's easier to work with rational functions, but if you can see (without reverse engineering) how to split the integral in the original form then there's no need to do this. $$I=\int_0^\frac{\pi}{4} \frac{\sin^2 x e^{-2mx}}{(m\sin x-\cos x)^2}dx\overset{\cot x \to x}=\int_1^\infty \frac{e^{-2m\cot^{-1}x}}{1+x^2}\frac{dx}{(m-x)^2}$$ Forcing a partial fraction to have $\frac{1}{1+x^2}$ and $\frac{1}{(m-x)^2}$ separately yields: $$I=\frac{1}{1+m^2}\int_1^\infty e^{-2m\cot^{-1} x}\left(\color{blue}{\frac{1}{1+x^2}\frac{m+x}{m-x}}+\color{red}{\frac{1}{(m-x)^2}}\right)dx$$

Finally integrating by parts the first term gives: $$\int_1^\infty \frac{e^{-2m\cot^{-1} x}}{\color{blue}{1+x^2}}\color{blue}{\frac{m+x}{m-x}}dx = \int_1^\infty \left(\frac{e^{-2m\cot^{-1} x}}{2m}\right)'\frac{m+x}{m-x}dx$$ $$=\frac{e^{-2m\cot^{-1} x}}{2m}\frac{m+x}{m-x}\bigg|_1^\infty -\int_1^\infty \frac{e^{-2m\cot^{-1} x}}{\color{red}{(m-x)^2}}dx$$ $$\Rightarrow I=\frac{1}{1+m^2}\frac{e^{-2m\cot^{-1} x}}{2m}\frac{m+x}{m-x}\bigg|_1^\infty=\frac{1}{2m(1+m^2)}\left(\frac{1+m}{1-m}e^{-m\pi/2}-1\right)$$

Solution 2:

Well the anti derivative exists. As pointed out by Claude Leibovici. I had just assumed that as it was a chapter on definite integrals, they would not give examples where the indefinite integral exists. Anyway here is a calculation of the anti derivative. The result on the last line.

\begin{equation*} \begin{split} \int e^{-2mx}\frac{\tan^2 x}{(1-m\tan x)^2}dx &=\int e^{-2mx}\frac{\sec^2 x}{(1-m\tan x)^2}dx -\int e^{-2mx}\frac{1}{(1-m\tan x)^2}dx\\ &=\frac{1}{m}\int e^{-2mx}d\frac{1}{(1-m\tan x)} -\int e^{-2mx}\frac{1}{(1-m\tan x)^2}dx\\ &=e^{-2mx}\frac{1}{m(1-m\tan x)}+2\int e^{-2mx}\frac{1}{(1-m\tan x)}dx\\ &-\int e^{-2mx}\frac{1}{(1-m\tan x)^2}dx\\ &=e^{-2mx}\frac{1}{m(1-m\tan x)}+2\int e^{-2mx}\frac{1-m\tan x}{(1-m\tan x)^2}dx\\ &-\int e^{-2mx}\frac{1}{(1-m\tan x)^2}dx\\ &=e^{-2mx}\frac{1}{m(1-m\tan x)}+\int e^{-2mx}\frac{1-2m\tan x}{(1-m\tan x)^2}dx\\ &=e^{-2mx}\frac{1}{m(1-m\tan x)}+\int e^{-2mx}\frac{1-2m\tan x+m^2\tan^2 x}{(1-m\tan x)^2}dx\\ &-m^2\int e^{-2mx}\frac{\tan^2 x}{(1-m\tan x)^2}dx \end{split} \end{equation*}

So $$(1+m^2)\int e^{-2mx}\frac{\tan^2 x}{(1-m\tan x)^2}dx= e^{-2mx}\frac{1}{m(1-m\tan x)}+\int e^{-2mx}dx$$ $$= e^{-2mx}\frac{1}{m(1-m\tan x)}-\frac{1}{2m}e^{-2mx}$$ $$= \frac{1}{2m}e^{-2mx}\frac{1+m\tan x}{1-m\tan x}$$

So finally,

$$\int e^{-2mx}\frac{\tan^2 x}{(1-m\tan x)^2}dx= \frac{1}{2m(1+m^2)}e^{-2mx}\frac{1+m\tan x}{1-m\tan x}$$