 Hi everyone. In this video, I want to show that the antiderivative tangent of x is equal to the natural log of the absolute value of secant of x plus a constant. Now, if you were given this, it wouldn't be too hard to show from what you're given right here, because we could just take the derivative of the natural log of the absolute value of secant of x right here, and just proceed and show that this is equal to tangent. But the issue that I want to, the way I want to talk about this one is, what if we didn't know what the derivative was, or what the antiderivative was? How could we come up with the antiderivative from scratch, right, and show that way, showing in that way what the antiderivative is? In that regard, we could use u substitution, because when you look at tangent here, one of the classic trigonometric identities is that tangent is sine over cosine. And sine and cosine have a nice kinship when it comes to derivatives, right? We know, for example, that the derivative of sine is equal to cosine, and we also know that the derivative of cosine is equal to negative sine. And so in some respects, sine and cosine are derivatives of each other up to a plus or minus, right? And so this process becomes reversible when it comes to antiderivatives. The antiderivative of sine dx is going to equal negative cosine of x plus a constant, and the antiderivative of cosine of x dx is equal to just sine of x dx. So you have this derivative antiderivative relationship with respect to cosine and sine. So in this situation, we have sine over cosine. Let's take the denominator to be our u, u equals cosine of x. And then as we can see there on the screen, the derivative of cosine is a negative sine of x dx. And so although we don't have a negative sine present, now we do. We can now take our antiderivative here, rewriting this in terms of the u substitution, our integral becomes negative the integral of du over u. And so notice here that the negative sine x dx becomes our entire du. And so this right here is just equal to du. And so we want to find an antiderivative negative antiderivative of one over u. And so recognizing the antiderivative of one over u is just the natural log there. You get negative the natural log of the absolute value of u plus a constant. In which case that's the same thing as negative the natural log of the absolute value of cosine of x dx. Because after all our u was cosine. And a lot of people are quite content with that as a final answer. But by properties of logarithms, a coefficient in front of them of the logarithm is equivalent to an exponent on the inside. And so raising something to the negative exponent takes its reciprocal. So we get the natural log of one over cosine of x and the reciprocal of cosine of course is just a secant. So we get the natural log of the absolute value of secant of x plus a constant. And so this then verifies the result we are looking for. Remember if we forgot what we wrote up here earlier. Oh, there it is. The antiderivative is natural log of the absolute value of secant of x plus C. So the thing here is this doesn't just give us the antiderivative of tangent. But by mimicking this strategy, we could also find the antiderivative of cotangent, right? And so I'll encourage the viewers of this video, right? If you want to be the first commenter, why don't you post what the antiderivative of cotangent is? Excuse me. Sort of mimicking this strategy right here.