Question Video: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions | Nagwa Question Video: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions | Nagwa

# Question Video: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions Mathematics • Higher Education

Find the function π, if πβ²(π‘) = 2 sec π‘(tan π‘ + 4 sec π‘), when βπ/2 < π‘ < π/2 and π(βπ/3) = β2.

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### Video Transcript

Find the function π, if π prime of π‘ equals two sec π‘ times tan π‘ plus four sec π‘, when π‘ is greater than negative π by two and less than π by two and π of negative π by three is negative two.

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

Now, this doesnβt look very nice. Letβs first distribute the parentheses of our integrand. When we do, we obtain our function π of π‘ to be equal to the integral of two sec π‘ tan π‘ plus eight sec squared π‘ with respect to π‘. 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 π of π‘ is equal to the integral of two sec π‘ tan π‘ with respect to π‘ plus the integral of eight sec squared π‘ with respect to π‘.

Now, these might look really nasty. But letβs recall some derivatives. We know that the derivative of sec of π‘ with respect to π‘ is tan π‘ sec of π‘. And we know that the derivative of tan π‘ with respect to π‘ is sec squared π‘. This means the antiderivative of tan of π‘ sec of π‘ must be sec of π‘. And the antiderivative of sec squared π‘ must be tan π‘. And so, the integral of two sec of π‘ tan π‘ must be two sec π‘. And, of course, itβs an indefinite integral. So weβll add a constant of integration π΄. Then the integral of eight sec squared π‘ must be eight tan π‘. And weβll add a second constant of integration π΅. Combining these two constants of integration, we see that π of π‘ is equal to two sec π‘ plus eight tan π‘ plus πΆ.

This is the general solution to our differential equation. But we havenβt yet used the fact that when π‘ is equal to negative π by three, our function is equal to negative two. So letβs substitute these values in. And when we do, we obtain negative two to be equal to two sec of negative π by three plus eight tan of negative π by three plus πΆ. If we recall that sec of negative π by three is one over cos of negative π by three. Then two sec of negative π by three must be two over cos of negative π by three, which gives us four. Then eight tan of negative π by three is negative eight root three. So negative two is equal to four minus eight root three plus πΆ.

Letβs subtract four from both sides of this equation and add eight root three to both sides. And we see that πΆ is equal to eight root three minus six. And weβve obtained the particular solution to our differential equation to be π of π‘ equals two sec π‘ plus eight tan π‘ plus eight root three minus six.

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