2
$\begingroup$

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

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

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

$u=\cos2x$

$du=-2\sin2x$

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

$\endgroup$
0

1 Answer 1

Reset to default
4
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ I was just a minute behind with the same path. $\endgroup$
    – Leucippus
    Feb 22, 2016 at 21:17
  • 1
    $\begingroup$ @Leucippus See math.stackexchange.com/questions/1194139/… ;-) $\endgroup$
    – egreg
    Feb 22, 2016 at 21:19
  • $\begingroup$ why in the denominator it is $\cos^2x$ and not $1-\cos^2x$? $\endgroup$
    – gbox
    Feb 22, 2016 at 21:30
  • 1
    $\begingroup$ @gbox Because $\cos2x+1=2\cos^2x-1+1=2\cos^2x$; in the numerator we have $\cos2x-1=1-2\sin^2x-1=-2\sin^2x$ $\endgroup$
    – egreg
    Feb 22, 2016 at 21:31
  • 1
    $\begingroup$ @gbox It's $[x-\tan x]_0^{\pi/4}=\frac{\pi}{4}-1$ $\endgroup$
    – egreg
    Feb 22, 2016 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.