 We find the integral of secant is nothing anyone would ever imagine. To find the integral of secant, we note that when evaluating an expression like this, we can use a trigonometric substitution to restate this as the integral of a trigonometric function. But if you can go forward, you can go backward, and so we can reverse the process and rewrite the integral of secant as a rational or radical expression. So suppose secant theta equals x, then dx equals secant theta tan theta, and so drawing our triangle tells us, and consequently the integral of secant is, which we can't evaluate. So remember, if the first thing you try doesn't work, try something else. So if we quickly review our trigonometric substitutions, we note that the substitution that involves secant typically uses something like an x squared minus a squared. So what if we had an integral with x squared minus a squared? So for example, how about this integral? Now we know we could do this using partial fractions to get. So while this does answer the question of what is the integral of one over x squared minus one, a good way to learn mathematics is to solve the same problem in different ways. So what if we use the substitution x equals secant theta? Then dx will be, and our integrand becomes, now since secant squared theta minus one equals tangent squared theta, we can simplify our integrand at rewriting tangent as sine divided by cosine, and secant as the reciprocal of cosine gives us. And we know the left-hand side because we can integrate it using partial fractions. And then since we use the substitution x equals secant theta, we get, and the secret to success is matching questions to answers. And so we find the integral of cosecant is. But we wanted to find secant of theta. So remember, how you speak influences how you think. And that leads us to the following. The co-functions are related. In particular, the cosine, cosecant, and co-tangent are the sine, secant, and tangent of the complementary angles. And likewise, the sine, secant, and tangent are the cosine, cosecant, and co-tangent of the complementary angles. So that means if we want to find the integral of secant, well, it's really the same as the integral of cosecant of the complementary angle, we can use a u-substitution, our integral of cosecant, put things back where you found them, and the secant of the complementary angle is the same as the cosecant of the angle. And so we have the integral of secant. And we can say that the integral of secant is equal to this thing. And a normal person would be happy that we solved the problem of finding the integral of secant. But mathematicians are not normal people. Sooner or later it comes down to style. Well this is true, it's an ugly mess, so let's try to simplify it. First of all, this negative log can be simplified. Negative log of A is the log of the reciprocal. And so we can invert our radical. Now when dealing with trigonometric functions, a difference of squares is always useful. So if we multiply by what we might think about as a conjugate, cosecant theta plus 1, we get, and since cotangent squared theta plus 1 equals cosecant squared theta, then we know that cosecant squared theta minus 1 is cotangent squared theta. And since we're taking the square root of a square, we get the absolute value. And we can simplify this a little bit further. Cosecant is the reciprocal of sine, and cotangent is cosine divided by sine. This gives us a compound fraction, so if we multiply numerator and denominator by sine theta, we can simplify and split. And reciprocal cosine is secant, and sine over cosine is tangent. And so we get our final result. The integral of secant is log secant plus tangent plus constant. Now we know that the derivative will be, which if we factor we can rewrite as. So we claim that we find the integral of secant by multiplying by this expression because it's obvious we want to do that. We'll use a u-substitution because this is the obvious u-substitution, and that gives us the log, and there's our integral. Yeah, right, sure, that's how you found it.