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#1

$Old$ January 11th, 2009, 07:55 AM

dillonmhudson $dillonmhudson is offline$

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$Default$ Integration problem?

How in the world do you solve this???:

int[sec(1-x)*tan(1-x)*dx]

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#2

$Old$ January 11th, 2009, 08:00 AM

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Hint

int(sec(x)tan(x))dx=sec(x)+c

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$Old$ January 11th, 2009, 08:05 AM

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Ok,

so using u-substitution:

u = sec(1-x)
du = sec(1-x)*tan(1-x)?

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#4

$Old$ January 11th, 2009, 08:33 AM

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Quote:

Originally Posted by dillonmhudson $View Post$

How in the world do you solve this???:

int[sec(1-x)*tan(1-x)*dx]

Thank you!

Remember that $\sec{(1 - x)} = \frac{1}{\cos{(1 - x)}}$ and that $\tan{(1 - x)} = \frac{\sin{(1-x)}}{\cos{(1 -x)}}$ .

Using these, we see that

$\int{\sec{(1-x)}\,\tan{(1-x)}\,dx} = \int{\frac{1}{\cos^2{(1-x)}}\,\sin{(1-x)}\,dx}$ .

Use the substitution $u = \cos{(1-x)}$ so that $\frac{du}{dx} = \sin{(1-x)}$ . Then the integral becomes

$\int{\frac{1}{u^2}\,\frac{du}{dx}\,dx}$

$= \int{u^{-2}\,du}$

$= -u^{-1} + C$

$= -\frac{1}{\cos{(1-x)}} + C$

$= -\sec{(1-x)} + C$ .

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#5

$Old$ January 11th, 2009, 08:36 AM

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$Default$ Awesome!

Thank you both very much!

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anti-differentiation, antiderivative, integrate, integration, trig

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