1. ## Integration problem?

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

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

Thank you!

2. ## Try it

Hint

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

3. Ok,

so using u-substitution:

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

Thanks!

4. Originally Posted by dillonmhudson
How in the world do you solve this???:

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

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

Using these, we see that

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

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

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

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

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

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

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

5. ## Awesome!

Thank you both very much!