How would I find the antiderivative of this equation?

S 0 to pi/3 (1-secxtans)dx

This should be one that you should just recognize from your study of trig derivatives.

Isn't the derivative of sec x, or (cos x)^-1
-(cos x)^-2(-sin x) ??
and isn't that sinx/cosx (1/cosx) ?
and isn't that tanxsecx ??

Can you take it from there?

To find the antiderivative of the equation S 0 to pi/3 (1-sec(x)tan(x)) dx, you can follow these steps:

Step 1: Use the trigonometric identity sec(x) = 1/cos(x) to rewrite the equation as S 0 to pi/3 (1 - (1/cos(x)) * (sin(x)/cos(x))) dx.

Step 2: Expand the parenthesis to get S 0 to pi/3 (1 - sin(x) / cos^2(x)) dx.

Step 3: Simplify the expression by multiplying the terms inside the integral by cos^2(x) to get S 0 to pi/3 (cos^2(x) - sin(x)) dx.

Step 4: Now you can integrate each term individually.

For the term cos^2(x), use the power rule for integration which states that the integral of x^n dx, where n is a constant not equal to -1, is (x^(n+1))/(n+1) + C. Therefore, the integral of cos^2(x) dx is (1/2) * (x + sin(2x)/2) + C.

For the term -sin(x), the integral is -cos(x) + C.

Step 5: Add the results of the integrals of each term to obtain the final antiderivative. Therefore, the antiderivative of (1 - sec(x)tan(x)) dx with limits 0 to pi/3 is [(1/2) * (x + sin(2x)/2) - cos(x)] evaluated from 0 to pi/3.

Evaluate the antiderivative at pi/3:

[(1/2) * (pi/3) + sin(2(pi/3))/2 - cos(pi/3)]

= (pi/6 + sqrt(3)/4 - 1/2)

Evaluate the antiderivative at 0:

[(1/2) * (0) + sin(2(0))/2 - cos(0)]

= (0 + 0 - 1)

Therefore, the value of the antiderivative of S 0 to pi/3 (1-sec(x)tan(x)) dx is (pi/6 + sqrt(3)/4 - 1/2) - (-1) which simplifies to (pi/6 + sqrt(3)/4 - 1/2) + 1.