Video: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions

Find the function 𝑓, if 𝑓′(𝑑) = 2 sec 𝑑(tan 𝑑 + 4 sec 𝑑), when βˆ’πœ‹/2 < 𝑑 < πœ‹/2 and 𝑓(βˆ’πœ‹/3) = βˆ’2.

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Video Transcript

Find the function 𝑓, if 𝑓 prime of 𝑑 equals two sec 𝑑 times tan 𝑑 plus four sec 𝑑, when 𝑑 is greater than negative πœ‹ by two and less than πœ‹ by two and 𝑓 of negative πœ‹ by three is negative two.

In this question, we have a differential equation. Remember, this is just an equation involving derivatives. Here, that’s 𝑓 prime of 𝑑. And we’re also given an initial value. That is, when 𝑑 is equal to negative πœ‹ by three, 𝑓 of 𝑑 is equal to negative two. Now, we’re looking to work out what the function 𝑓 is. And so, we recall that integration and differentiation are reverse processes. So our original function 𝑓 of 𝑑 will be the integral of 𝑓 prime of 𝑑 with respect to 𝑑. That’s the integral of two sec of 𝑑 times tan 𝑑 plus four sec of 𝑑 with respect to 𝑑.

Now, this doesn’t look very nice. Let’s first distribute the parentheses of our integrand. When we do, we obtain our function 𝑓 of 𝑑 to be equal to the integral of two sec 𝑑 tan 𝑑 plus eight sec squared 𝑑 with respect to 𝑑. We can simplify this a little bit by recalling that the integral of the sum of two functions is equal to the sum of the integral of each of those functions. So 𝑓 of 𝑑 is equal to the integral of two sec 𝑑 tan 𝑑 with respect to 𝑑 plus the integral of eight sec squared 𝑑 with respect to 𝑑.

Now, these might look really nasty. But let’s recall some derivatives. We know that the derivative of sec of 𝑑 with respect to 𝑑 is tan 𝑑 sec of 𝑑. And we know that the derivative of tan 𝑑 with respect to 𝑑 is sec squared 𝑑. This means the antiderivative of tan of 𝑑 sec of 𝑑 must be sec of 𝑑. And the antiderivative of sec squared 𝑑 must be tan 𝑑. And so, the integral of two sec of 𝑑 tan 𝑑 must be two sec 𝑑. And, of course, it’s an indefinite integral. So we’ll add a constant of integration 𝐴. Then the integral of eight sec squared 𝑑 must be eight tan 𝑑. And we’ll add a second constant of integration 𝐡. Combining these two constants of integration, we see that 𝑓 of 𝑑 is equal to two sec 𝑑 plus eight tan 𝑑 plus 𝐢.

This is the general solution to our differential equation. But we haven’t yet used the fact that when 𝑑 is equal to negative πœ‹ by three, our function is equal to negative two. So let’s substitute these values in. And when we do, we obtain negative two to be equal to two sec of negative πœ‹ by three plus eight tan of negative πœ‹ by three plus 𝐢. If we recall that sec of negative πœ‹ by three is one over cos of negative πœ‹ by three. Then two sec of negative πœ‹ by three must be two over cos of negative πœ‹ by three, which gives us four. Then eight tan of negative πœ‹ by three is negative eight root three. So negative two is equal to four minus eight root three plus 𝐢.

Let’s subtract four from both sides of this equation and add eight root three to both sides. And we see that 𝐢 is equal to eight root three minus six. And we’ve obtained the particular solution to our differential equation to be 𝑓 of 𝑑 equals two sec 𝑑 plus eight tan 𝑑 plus eight root three minus six.

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