What is #f(x) = int secx-cotx dx# if #f(pi/8) = 0 #?

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#

Now, we must find the integrals of each part. For the first one, we will multiply by #(sec x+tan x)# on both the top and bottom. We get away with this because #(secx+tanx)/(secx+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)#

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/y# situation

#d/dx (secx+tanx) = d/dx (1/cosx + (sinx)/cosx) = (sinx)/cos^2x + (cos^2x+sin^2x)/(cos^2x)= secxtanx+ sec^2x#

Thus, we have a #(du)/u# situation, with #u = secx+tanx, du = sec^2x+secxtanx dx#

The integral of such an expression is simply #ln|u| + c = ln|secx+tanx|+c#

To integrate our second part, the #-cotx#, we work similarly, recalling that #cotx = cosx/sinx#...

#-intcotxdx = -int cosx/sinx dx#

Here, we already have a #(du)/u#. We know from above then that we will have an answer of the form #lnu+c#, but we must not forget the - sign.

#= -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#

To get rid of the constant, we plug in the term #pi/8#. According to calculations, that yields us:

#ln|sec(pi/8) + tan(pi/8)| - ln |sin (pi/8)| + c = 0 approx 1.364+c#

Thus, #c = -1.364#, and...

#f(pi/8)-0, int(secx-cotx)dx approx ln|secx+tanx|-ln|sinx|-1.364#