To find this integral, we will have to rely on some trigonometric tricks.
First, separate the integral into 2 parts:
int(secx-cotx)dx=intsecxdx-intcotxdx∫(secx−cotx)dx=∫secxdx−∫cotxdx
Now, we must find the integrals of each part. For the first one, we will multiply by (sec x+tan x)(secx+tanx) on both the top and bottom. We get away with this because (secx+tanx)/(secx+tanx)=1secx+tanxsecx+tanx=1, so we're simply multiplying by a form of 1.
int secxdx = int secx(secx+tanx)/(secx+tanx) = int (sec^2x+secxtanx)/(secx+tanx)∫secxdx=∫secxsecx+tanxsecx+tanx=∫sec2x+secxtanxsecx+tanx
It seems we will use a natural log here, but it helps for us to confirm it.
Differentiate the denominator; if it turns out equal to the numerator, we have a dy/ydyy situation
d/dx (secx+tanx) = d/dx (1/cosx + (sinx)/cosx) = (sinx)/cos^2x + (cos^2x+sin^2x)/(cos^2x)= secxtanx+ sec^2xddx(secx+tanx)=ddx(1cosx+sinxcosx)=sinxcos2x+cos2x+sin2xcos2x=secxtanx+sec2x
Thus, we have a (du)/uduu situation, with u = secx+tanx, du = sec^2x+secxtanx dxu=secx+tanx,du=sec2x+secxtanxdx
The integral of such an expression is simply ln|u| + c = ln|secx+tanx|+cln|u|+c=ln|secx+tanx|+c
To integrate our second part, the -cotx−cotx, we work similarly, recalling that cotx = cosx/sinxcotx=cosxsinx...
-intcotxdx = -int cosx/sinx dx−∫cotxdx=−∫cosxsinxdx
Here, we already have a (du)/uduu. We know from above then that we will have an answer of the form lnu+clnu+c, but we must not forget the - sign.
= -ln|sinx|-c=−ln|sinx|−c
Adding these together, and remembering that c is any constant, we get:
int(secx-cotx)dx = ln|secx+tanx| - ln |sinx|+c∫(secx−cotx)dx=ln|secx+tanx|−ln|sinx|+c
To get rid of the constant, we plug in the term pi/8π8. According to calculations, that yields us:
ln|sec(pi/8) + tan(pi/8)| - ln |sin (pi/8)| + c = 0 approx 1.364+cln∣∣sec(π8)+tan(π8)∣∣−ln∣∣sin(π8)∣∣+c=0≈1.364+c
Thus, c = -1.364c=−1.364, and...
f(pi/8)-0, int(secx-cotx)dx approx ln|secx+tanx|-ln|sinx|-1.364f(π8)−0,∫(secx−cotx)dx≈ln|secx+tanx|−ln|sinx|−1.364