Question

Find the integral. Use a computer algebra system to confirm your result. $$ \int \frac{1-\sec t}{\cos t-1} d t $$

Short answer

The integral of the given function is \[- \ln |cos(t) - 1| + C.\]This result can be verified using a computer algebra system.

Step-by-step solution

Simplify the integrand using trigonometric identities

Start by simplifying the integral using the identity \(\sec(t) = 1/\cos(t)\). So, the given integral becomes \[ \int \frac{1- 1/cos(t)}{cos(t)-1} dt.\] We can simplify it further by multiplying the numerator and denominator by \(cos(t)\), resulting in \[ \int \frac{cos(t)-1}{cos^2(t) - cos(t) - 1} dt.\]

Implement an appropriate substitution

Let's perform a substitution, where \(cos(t) = x + 1\). Its differential \(dt = -dx/sin(t)\). By squaring on both sides, we have \(cos^2(t) = (x + 1)^2, which simplifies the integral as \[ \int \frac{x}{(x+1)^2 - (x+1) - 1}\] multiplied by \(-dx\).

Simplify and calculate the integral

The final simplification gives us: \[ -\int \frac{x}{x^2} dx\] which is a simple polynomial term and can be integrated using basic integration rules to give: \[-\int x^{-1} dx = - \ln |x| + C\]

Undo the substitution

Now, substitute \(x\) back in terms of \(t\), \(x = cos(t) - 1\). So the final answer is: \[- \ln |cos(t) - 1| + C.\]