Question Video: Finding the General Antiderivative of a Given Function Involving Trigonometric and Exponential Functions Mathematics • Higher Education

Determine ∫ βˆ’3𝑒^πœƒ + 2 tan πœƒ sec πœƒ π‘‘πœƒ.

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Video Transcript

Determine the integral of negative three 𝑒 to the πœƒ plus two tan πœƒ sec πœƒ π‘‘πœƒ.

In this question, we are asked to integrate an expression with respect to πœƒ. To do this, we will split the integrand so we are integrating term by term. We will begin by integrating negative three 𝑒 to the πœƒ with respect to πœƒ. We recall that the integral of 𝑒 to the power of π‘₯ with respect to π‘₯ is simply 𝑒 to the power of π‘₯ plus a constant of integration 𝑐. This means that the integral of negative three 𝑒 to the power of πœƒ is simply negative three 𝑒 to the power of πœƒ. At this stage, we will not add a constant of integration as we can just do this at the end of our expression.

Next, we need to consider the integral of two tan πœƒ sec πœƒ. One way of doing this is to first recall that the derivative of sec πœƒ with respect to πœƒ is tan πœƒ multiplied by sec πœƒ. This tells us that sec πœƒ is an antiderivative of tan πœƒ sec πœƒ. And hence, the integral of tan πœƒ sec πœƒ with respect to πœƒ is equal to sec πœƒ plus a constant of integration 𝑐.

We notice that this expression appears in our integrand. However, it has been multiplied by two. The integral of two tan πœƒ sec πœƒ is therefore equal to two sec πœƒ. Adding a constant of integration 𝑐, we now have our final answer. The integral of negative three 𝑒 to the πœƒ plus two tan πœƒ sec πœƒ with respect to πœƒ is equal to negative three 𝑒 to the πœƒ plus two sec πœƒ plus 𝑐.

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