How to approach this integral besides Tangent half-angle substitution $\int \frac{dx}{\sin x-\cos x}$

Question

Is there any faster method to solve this integral besides tangent half-angle substitution

$$\int \frac{dx}{\sin x-\cos x}$$

I was trying

$$\int \frac{\sin^2 x+\cos^2 x}{\sin x-\cos x}$$ $$\int \frac{\sin^2 x}{\sin x-\cos x}dx + \int \frac{\cos^2 x}{\sin x-\cos x}dx$$

score 1 · Accepted Answer · answered Dec 02 '23 at 22:52

1

HINT

I would recommend you to notice that:

\begin{align*} \sin(x) - \cos(x) & = \sqrt{2}\left(\sin(x)\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\cos(x)\right)\\\\ & = \sqrt{2}\left(\sin(x)\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\cos(x)\right)\\\\ & = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right) \end{align*}

Can you proceed from here?

answered Dec 02 '23 at 22:52

Átila Correia

15,530

How did you come up with this idea? – samsamradas Dec 02 '23 at 22:54
@samsamradas It's a standard technique. – J.G. Dec 02 '23 at 22:57
@Átila It was a duplicate https://math.stackexchange.com/questions/3407227 – Anne Bauval Dec 02 '23 at 22:58
@AnneBauval thanks for sharing. I shall propose a slightly different approach from the given reference. – Átila Correia Dec 02 '23 at 23:00
If so, shouldn't you rather do it over there? (and vote to close this one as a duplicate and delete your answer here) – Anne Bauval Dec 02 '23 at 23:02
@ÁtilaCorreia should I take the natural log of the sin to solve the integral or rather use csc – samsamradas Dec 02 '23 at 23:03
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@AnneBauval Thanks for the suggestion. I will do it there. – Átila Correia Dec 02 '23 at 23:04
@samsamradas make the substitution $u = x - \frac{\pi}{4}$ and integrate the $\csc$ function as you mentioned. – Átila Correia Dec 02 '23 at 23:08
@ÁtilaCorreia why $ln|sin(u)|$ is not valid? – samsamradas Dec 02 '23 at 23:10
@samsamradas because $(\ln|f|)'=f'/f,$ not $1/f.$ But see https://math.stackexchange.com/questions/331307 – Anne Bauval Dec 02 '23 at 23:10
@AnneBauval thanks – samsamradas Dec 02 '23 at 23:12

How to approach this integral besides Tangent half-angle substitution $\int \frac{dx}{\sin x-\cos x}$

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