Question Video: Finding the Integration of a Function Involving Trigonometric Functions Using Integration by Substitution Mathematics

Video Transcript

Determine the integral of negative seven tan 𝑥 sec to the power of four 𝑥.

Well, the first thing I’ve done is that I’ve written it the other way around. So I just rearranged it, so we’ve got the integral of negative seven sec to power of four 𝑥 tan 𝑥. So what’s the next step? So the next step is to use a substitution. The substitution I’ve chosen to use is 𝑢 is equal to sec to the power of four 𝑥. And then what we do is we find d𝑢 d𝑥. So d𝑢 d𝑥 is four sec to the power of four 𝑥 tan 𝑥. And we got that by differentiating sec to the power of four 𝑥.

So we’re just gonna quickly recap how we got to that result. Well first of all, one of the things we know which is a general solution that we have is that if you differentiate sec 𝑥 you get sec 𝑥 tan 𝑥. So now, let’s look at the chain rule because a chain rule is actually what you would use to differentiate sec to the power of four 𝑥. So if you know that you’ve got a function 𝑦 is equal to 𝑓 of 𝑢, then d𝑦 d𝑥 is equal to d𝑦 d𝑢 multiplied by d𝑢 d𝑥.

So in our case, our 𝑢 would be sec 𝑥. And our 𝑦 would be 𝑢 to the power of four, so therefore d𝑢 d𝑥. So our derivative of 𝑢 is gonna be sec 𝑥 tan 𝑥 because that’s one of our standard derivatives. And d𝑦 d𝑢 is gonna be equal to four 𝑢 cubed. So now to find d𝑦 d𝑥, what we’re gonna do is multiply these two results together. So we get four 𝑢 cubed multiplied by sec 𝑥 tan 𝑥. So then when we substitute in 𝑢 equal sec 𝑥, we get d𝑦 d𝑥 is equal to four sec cubed 𝑥 multiplied by sec 𝑥 multiplied by tan 𝑥 which is gonna give us our four sec to the power of four 𝑥 tan 𝑥. And we got that because if you have sec cubed 𝑥 multiplied by sec 𝑥, we get sec to power of four 𝑥.

Okay, great! So I’ve just shown you there how we would’ve come to our derivative. So therefore the 𝑥 is equal to one over four sec to the power of 𝑥 times tan 𝑥 d𝑢. So how we’re gonna use this? So what we can use this for is to substitute it back into our original integral. So what I’ve done here is I’ve taken the constants outside of the integration. So we’ve got negative seven over four. And then what we’ve got is sec to the power of four 𝑥 tan 𝑥 over sec to the power of four tan 𝑥 d𝑢. Well, these will cancel themselves out because we can divide top and bottom just by sec to the power of four 𝑥 tan 𝑥.

And what we’re left with is negative seven over four multiplied by the integral of one which will just be equal to negative seven 𝑢 over four and that’s because if we integrate one, we’re just gonna get 𝑢 because we’re doing it looking at 𝑢. So we’ve got negative seven 𝑢 over four. So then all we need to do is substitute back in what our 𝑢 value is. And we had 𝑢 is equal to sec to the power of four 𝑥.

And when we do that, we get negative seven sec to the power of four 𝑥 over four plus 𝐶 which could also be written as negative seven over four sec to the power of four 𝑥 plus 𝐶. So great, what we’ve done is we’ve determined the integral of negative seven tan 𝑥 sec to the power of four 𝑥. And that’s by using substitution.