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

1 Answer
Nov 23, 2017

Use trigonometric identities, u-substitution, and the known derivative formulae for various trig functions. f(x) approxln|secx+tanx|-ln|sinx|-1.364

Explanation:

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