Question

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

Answer 1

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`