Video Transcript
Determine the integral of two π₯ plus the sec squared of six π₯ with respect to π₯.
In this question, weβre asked to evaluate the integral of the sum of two functions, and we know how to integrate each of these two terms separately. So, the first thing weβll do is split our integral into two. Weβll split it into the integral of two π₯ with respect to π₯ plus the integral of the sec squared of six π₯ with respect to π₯. And we can evaluate each of these integrals directly. Since our first term is a linear function, we can integrate this by using the power rule for integration.
We want to add one to our exponent of π₯ and then divide by this new exponent. And to do this, we need to recall that π₯ is equal to π₯ to the first power. So, our exponent of π₯ is equal to one. So, weβll add one to our exponent of one, giving us a new exponent of two, and then divide by this new exponent, giving us two π₯ squared divided by two. And itβs worth pointing out here we donβt need to add our constant of integration because weβll get another constant of integration in our second integral. And we can just combine these two constants into one. And we can simplify two π₯ squared divided by two to give us π₯ squared.
Now, we want to evaluate the integral of the sec squared of six π₯ with respect to π₯. And thereβs two different ways we could do this. We could recall the derivative of the tan of six π₯ with respect to π₯ is six times the sec squared of six π₯. This would then tell us the tan of six π₯ all divided by six is an antiderivative of our integral. However, we already did this in the general case. We found if π is not equal to zero, then the integral of the sec squared of ππ₯ with respect to π₯ is equal to the tan of ππ₯ divided by π plus our constant of integration πΆ. This is a standard trigonometric integral result which comes up a lot. Itβs worth committing this to memory.
In our case, the value of π is equal to six. So, we set our value of π equal to six. And weβll write the result as one six times the tangent of six π₯ plus our constant of integration πΆ. And this gives us our final answer.
Therefore, we were able to show the integral of two π₯ plus the sec squared of six π₯ with respect to π₯ is equal to π₯ squared plus one-sixth times the tan of six π₯ plus our constant of integration πΆ.