# Integration problem?

• Jan 11th 2009, 05:55 AM
dillonmhudson
Integration problem?
How in the world do you solve this???:

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

Thank you!
• Jan 11th 2009, 06:00 AM
nikhil
Try it
Hint

int(sec(x)tan(x))dx=sec(x)+c
• Jan 11th 2009, 06:05 AM
dillonmhudson
Ok,

so using u-substitution:

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

Thanks!
• Jan 11th 2009, 06:33 AM
Prove It
Quote:

Originally Posted by dillonmhudson
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

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

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

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

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

$= -\sec{(1-x)} + C$.
• Jan 11th 2009, 06:36 AM
dillonmhudson
Awesome!
Thank you both very much!