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what am I doing wrong? math

I'm doing integration techniques and I don't know what I'm doing wrong here...

∫arcsec(2x) dx (x > 1/2)

Let u = arcsec(2x), du = 2/x*sq.rt(x^2 - 1), dv = dx, v = x

∫arcsec(2x) = xarcsec(2x) - 2∫1/sq.rt(x^2 - 1) dx
I cancelled the x's (x in the denominator of du is just x since x is always a positive value)

Now, using (x^2 - 1) with reference to a right triangle, x is the hypotenuse, and the measurement of the legs are 1 and sq.rt(x^2 - 1). I decided to use 1 as the bottom leg and sq.rt(x^2 - 1) as the right leg where theta is an acute positive angle.

tan(θ) = sq.rt(x^2 - 1)
sec(θ) = x which means sec(θ)tan(θ) dθ = dx

Rewrote integral as

∫arcsec(2x) = xarcsec(2x) - 2∫(sec(θ)tan(θ))/tan(θ)) dθ

The tangents cancel and the integral of sec(θ) comes out to be 2ln|sec(θ) + tan(θ)| so substituting back from θ -> x

F(x) = xarcsec(2x) - 2ln|x + sq.rt(x^2 + 1)| + C

but this is wrong. Anyone have any ideas? I can't see where my error is.

September 7, 2012

9 Comments • Newest first

DrHye

[quote=PyroWizard45]When do you start learning this stuff?[/quote]

Depends on a number of factors. You could take it as early as high school sophomore year (Calc BC), but most take it in college (Calc II). Usually anyone in college that is required to take calc II takes it within the first year of college, because it's a prerequisite or corequisite to other classes.

Reply September 7, 2012
ThatBox

...
Gee... I can't wait to start learning this...

Reply September 7, 2012
Oyster

I think BC calc.

Reply September 7, 2012
Chema

[quote=PyroWizard45]When do you start learning this stuff?[/quote]
Depends on which career you choose

Reply September 7, 2012
PyroWizard45

When do you start learning this stuff?

Reply September 7, 2012
Chema

[quote=shananay]What is this sorcery?[/quote]
Integrals, a delicious headache

Reply September 7, 2012
Suryoyo

2+2=4/2=2

Reply September 7, 2012
SunsetDews

[quote=Chema]Integrate by parts, then use a trig substitution.

u = arcsec(2x), du = du/[x√(4x² - 1)], dv = dx, v = x

�' arcsec 2x dx = x arcsec(x) - �' 1/√(4x² - 1) dx

Then, let x = ½secΘ, du = ½secΘtanΘ dΘ

�' 1/√(4x² - 1) dx = ½ �' secΘ dΘ = ½ln|secΘ + tanΘ| + C.

�' arcsec 2x dx = x arcsec(2x) - ½ ln|2x + √(4x² - 1)| + C[/quote]

Oh wow I'm being an idiot, I completely messed up the beginning.

Thanks.

Reply September 7, 2012
Chema

Integrate by parts, then use a trig substitution.

u = arcsec(2x), du = du/[x√(4x² - 1)], dv = dx, v = x

�' arcsec 2x dx = x arcsec(x) - �' 1/√(4x² - 1) dx

Then, let x = ½secΘ, du = ½secΘtanΘ dΘ

�' 1/√(4x² - 1) dx = ½ �' secΘ dΘ = ½ln|secΘ + tanΘ| + C.

�' arcsec 2x dx = x arcsec(2x) - ½ ln|2x + √(4x² - 1)| + C

Reply September 7, 2012 - edited