 So I want to talk some more about trigonometric integrals, but I want to change the direction a little bit we've done lots of examples focusing on combinations of signs and cosines and I think we've basically done all the ones we need to that is to say that any other Examples you'll run across like in the homework or in exams or wherever else If you've been practicing these ones so far you have the skill set necessary to handle those ones as well So I want to switch over to trigonometric functions that use tangents and secants And so before we jump into them Let me remind you of some of the important identities one should know involving tangents and secants The reason tangent and secant become such good friends is first of all the derivative of tangent is secant squared, right? The derivative of secant is tangent and secant So notice taking the derivatives of tangent and secant involves tangents and secants themselves This is useful observation when one tries to do a u-substitution and then also the Pythagorean relationship connects tangent and secants Together tangent squared plus one is equal to secant squared One can see this identity by taking the standard sine squared plus cosine squared equals one and dividing everything by cosine squared All right, so How does one find an anti derivative of an integral using sine or using tangents and secants? Well, one possibility is you could switch everything to sines and cosines. That's always a possibility I take a lot to students in basic trigonometry if you're trying to prove identities. It works here as well Tangent squared does become a tangent the sixth excuse me comes as sine to the sixth over cosine to the sixth And then you also get this one over cosine To the fourth x dx you could combine that together take sine to the sixth over cosine to the tenth And try to do something with that. I would say that that is a much more difficult approach, although doable I the fact that we have some secants and tangents here is actually our advantage and so let's try that right here So when we worked with sines and cosines in the past the idea is is there some way we could utilize a u-substitution right and so if we're going to use a u-substitution We want to look for specific to use the derivatives of functions like secant squared our tangent and secant Now when you have in your integral looking at this right here if the power of secant is an even number which uh-huh We have a four right there. If you have an even number of secants My recommendation is to use this identity right here and set this equal to du that as you want du to Equal secant squared dx now if du is equal to secant squared that means you should equal tangent and So let's see what we have right here. We do have six Tangents that right there would equate to be a u to the sixth What about the secant to the fourth? Well, it breaks up into two pieces. There's a secant squared x times a secant squared x Now the first secant are well the first or second doesn't matter one of the secant squareds We're gonna set aside as our du, but what do we do with the other secant squared? It can't be a du as well because then we'd have like a du squared. No, we don't want that Well, that's where the Pythagorean identity comes into play secant squared is the same thing as a tangent squared plus one So we can convert all the secants into tangents and so if we see what happens there Like I said, you're tangent to the sixth will become a u to the sixth Then one of the secant squareds will become a u squared plus one It's a tangent square plus one but remember tangent is you right now and then the other secant squared becomes a du So we're able to convert this trigonometric function into a polynomial function using this u substitution Distributing the u to the sixth throughout here. We have a very nice polynomial u to the eighth plus u to the sixth du and From here using the power rule for anti derivatives We can very easily find out the anti derivative will be one ninth u to the nine plus one seventh u to the seven plus a constant and then remembering that you itself is Just tangent of x we substitute that back in to get one ninth Tangent to the ninth x plus one seventh Tangent to the seventh x plus a constant and then this gives us our Anti-derivative right here and so I want you to mimic this strategy on future examples If you're trying to integrate a product of tangent and secant and you have an even number of secants Take u to be tangent and du to be secant squared and you should probably be just fine