# integration: partial fractions question (1 Viewer)

#### googleplex

##### New Member
integ x secx tanx dx

I got:
x secx - ln|tan(45 + x/2)| + C

45 being in degrees

I'm pretty sure it's right, but am not sure how the book got
x secx - ln(secx + tanx) + C

Is my answer wrong? How did they get theres?

Thanks

#### spice girl

##### magic mirror
tan(45 + x/2) = [1 + tan(x/2)] / [1 - tan(x/2)]
= [1 + tan(x/2)] ^2 / 1 - tan^2(x/2)
= [1 + 2tan(x/2) + tan^2(x/2)] / 1 - tan^2(x/2) = *

let t = tan(x/2)

then * = 1 + 2t + t^2 / 1 - t^2
from t-results, we know that 2t/1 - t^2 = tanx, and 1 + t^2 / 1 - t^2 = 1/cosx = secx

so tan(45 + x/2) = secx + tanx

#### googleplex

##### New Member
= [1 + tan(x/2)] ^2 / 1 - tan^2(x/2)
I'm totally lost. how did u just square the top and then square the tan on the bottom?

#### spice girl

##### magic mirror
Originally posted by googleplex
I'm totally lost. how did u just square the top and then square the tan on the bottom?
multiply top and bottom by 1 + tan(x/2). i jumped a step and expanded the denominator already.

#### onepac

##### New Member
integrate xsecxtanx by parts by letting u=x, v'=secxtanx
u'=1, v=secx
thus
integration of xsecxtanx.dx = xsecx - integration of secx.dx

= xsecx - log(secx+tanx) + C

Last edited:

#### Richard Lee

##### Member
Origin = int. xdsecx = x secx - int. secxdx = x secx - ln|secx + tanx| + C
Good luck!

#### KeypadSDM

##### B4nn3d
Just as an aside, never use degrees in integration, ever, ever, ever. Ever.

And again, ever.

Teachers go CRAZY, ABSOLUTELY CRAZY.

Plus it's technically wrong, d/dxSin[x] is only Cos[x] iff x is in radians.