integrate $\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$

Question

$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$$

$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}\cdot\frac{\cos 2x-1}{\cos 2x-1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos^2 2x-2\cos2x+1}{\cos^22x-1}dx$$

$$=\int^{\frac{\pi}{4}}_{0} \frac{\cos^2 2x-2\cos2x+1}{-\sin^22x}dx$$

$u=\cos2x$

$du=-2\sin2x$

Is this substitution is ok? or do dx must be in the numerator?

egreg · Accepted Answer · 2016-02-22T21:21:34.797

4

Remember the bisection/duplication formulas and note that $$\cos2x=2\cos^2x-1=1-2\sin^2x,$$ so your integral is $$ \int_{0}^{\pi/4}-\frac{\sin^2x}{\cos^2x}\,dx = \int_{0}^{\pi/4}\frac{\cos^2x-1}{\cos^2x}\,dx = \int_{0}^{\pi/4}\left(1-\frac{1}{\cos^2x}\right)\,dx $$ Can you finish?

edited Feb 22 '16 at 21:21

answered Feb 22 '16 at 21:16

egreg

244,946

I was just a minute behind with the same path. – Leucippus Feb 22 '16 at 21:17
1

@Leucippus See http://math.stackexchange.com/questions/1194139/is-there-another-way-to-solve-this-integral ;-) – egreg Feb 22 '16 at 21:19
why in the denominator it is $\cos^2x$ and not $1-\cos^2x$? – gbox Feb 22 '16 at 21:30
1

@gbox Because $\cos2x+1=2\cos^2x-1+1=2\cos^2x$; in the numerator we have $\cos2x-1=1-2\sin^2x-1=-2\sin^2x$ – egreg Feb 22 '16 at 21:31
@egreg can I write $x-tanx$ from 0 to $\frac{\pi}{4}$ or should it be xfrom 0 to $\frac{\pi}{4}$ -\anx from 0 to $\frac{\pi}{4}$ or it does not matter? – gbox Feb 22 '16 at 21:42
1

@gbox It's $[x-\tan x]_0^{\pi/4}=\frac{\pi}{4}-1$ – egreg Feb 22 '16 at 21:46

integrate $\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$

1 Answers1