3
$\begingroup$

I came across the following indefinite integral $$ \int \frac{2\,dx}{(\cos(x) - \sin(x))^2} $$ and was able to solve it by doing the following: First I wrote $$\begin{align*} \int \frac{2\,dx}{(\cos(x) - \sin(x))^2} &= \int \frac{2\,dx}{1 - 2\cos(x)\sin(x)} \\ &= \int \frac{2\,dx}{1 - \sin(2x)} \\ \end{align*}$$

Then setting $2x = z$, $$\begin{align*} &= \int \frac{dz}{1-\sin(z)}\\ &= \int \frac{1+\sin(z)}{\cos^2(z)}\,dz \\ &= \int \sec^2(z) + \tan(z)\sec(z) \,dz \\ &= \tan(z) + \sec(z) \\ &= \tan(2x) + \sec(2x). \end{align*}$$

The solutions to the problem were given as $\tan(x + \pi/2)$ or $\frac{\cos(x) + \sin(x)}{\cos(x) - \sin(x)}$. I checked that these solutions are in face equivalent to my solution of $\tan(2x) + \sec(2x)$.

My question is, are there other ways to calculate this integral that more "directly" produce these solutions? Actually, any elegant calculation methods in general would be interesting.

$\endgroup$

4 Answers 4

Reset to default
5
$\begingroup$

$$\int\frac{dx}{(\cos x-\sin x)^2}=\int\frac{dx}{(\sqrt2\cos(x+\frac\pi4))^2}$$ where you recognize the derivative of a tangent.

A very general and useful method is the use of the exponential representation

$$\cos x:=\frac{e^{ix}+e^{-ix}}2,\\\sin x:=\frac{e^{ix}-e^{-ix}}{2i}$$ together with the change of variable $z:=e^{ix}$ such that $dx=dz/iz$.

In your case

$$\int\frac{dx}{(\cos x-\sin x)^2} =\int \frac{4\,dz}{(z+z^{-1}+iz-iz^{-1})^2iz} =\int \frac{2\,d(z^2)}{((1+i)z^2+(1-i))^2i}$$ which is elementary.

$\endgroup$
5
$\begingroup$

HINT:

$$1-\sin2x=1-\cos2\left(\dfrac\pi4-x\right)=2\sin^2\left(\dfrac\pi4-x\right)=\dfrac2{\csc^2\left(\dfrac\pi4-x\right)}$$

As $\csc(-A)=-\csc A,$

$$\csc^2\left(\dfrac\pi4-x\right)=\csc^2\left(x-\dfrac\pi4\right)$$

$$\int\csc^2y\ dy=-\cot y+K$$

Alternatively,

$$1-\sin2x=\dfrac{(1-\tan x)^2}{1+\tan^2x}$$

$\endgroup$
5
$\begingroup$

see my nice answer:

$$\int \frac{2\ dx}{(\cos x-\sin x)^2}=\int \frac{2\ dx}{\cos^2 x\left(1-\frac{\sin x}{\cos x}\right)^2}$$ $$=2\int \frac{\sec^2 x\ dx}{\left(1-\tan x\right)^2}$$ $$=-2\int \frac{d(1-\tan x)}{\left(1-\tan x\right)^2}$$ $$=-2 \frac{-1}{\left(1-\tan x\right)}+C$$$$=\frac{2\cos x}{\cos x-\sin x}+C$$

$\endgroup$
0
$\begingroup$

We are going to evaluate the integral by auxiliary angle. $$ \begin{aligned} \int \frac{2 d x}{(\cos x-\sin x)^{2}} &=\int \frac{2 d x}{\left[\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)\right]^{2}} \\ &=\int \sec ^{2}\left(x+\frac{\pi}{4}\right) d x \\ &=\tan \left(x+\frac{\pi}{4}\right)+C \\ (\textrm{ OR }&=\frac{2 \sin x}{\cos x-\sin x}+C’) \end{aligned} $$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.