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Evaluate the indefinite integral. Illustrate, and…

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Problem 53 Hard Difficulty

Evaluate the indefinite integral. Illustrate, and check that your answer is reasonable, by graphing both the integrand and its antiderivative (taking $ C = 0 $).

$ \displaystyle \int \sin 3x \sin 6x dx $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

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Video Transcript

no For this problem will find the anti derivative of signed three X time science XX Then we'LL check. Their answer is reasonable by first graphing the interbrand which is circled in red as well as the entire anti derivative. So first, let's just go ahead and find the anti derivatives. So one way to proceed is to go ahead and re write us. I noticed that six X is two times three x So I take the second determine here and I write is sign of two times three x. So this is corresponding to the science six X Here I did was rewrite this No into this. And then from this new term sign of two times three x, we can use this double angle formula to rewrite this. So we have the integral signed three eggs. Times go different color here, So we have to signed three eggs. Times co signed three X. So here we're using T equals three X and our problem. Okay, so then we could go ahead and simplify than integrate. So here is pulled. That too. We have a sine squared three x cosign three x and then this looks like we should do it. Use up. Let's take you two be signed three x then do you is coastline three eggs times three by the general So that do you over three is co signed three x So here we have to theirs in a girl You square Do you which is to you cute over nine plus c by the instructions were supposed to take seed of zero so we could go in and drop this constant here and then lastly, we can go back to our U substitution toe back substitute. So here we have to over nine Sign Cube, three x. So this is our final answer. This is our anti derivatives. So let's call this in blue. No. So and again we've coordinated the instagram with red. So if the integral of the red function equals the blue function, that means that the derivative of the blue function so derivative of the blue equals the red. Then we can go to the graphing calculator toe check. If this is reasonable, so score the gravity calculator. So the red graph is supposed to be the derivative of the blue graph. So let's first check to see when the red graph should be zero. So let's see, when the blue graph has a horizontal tangent. Here at the origin, we see a horizontal tension that means the rivet of a zero and the Red graphic zero. Here, at about point five, we see a horizontal tangent on the blue. That means that the riveting zero and that's where the red is. And same over here. At about one point one, we see a horizontal changes on the blue and the road graphic zero, and you could go out and check that every time. The blue graph has a horizontal tension, the red graph zero and maybe from to convince you further hear from zero to about point five, we see that the blue graph is increasing. That means that the derivative is positive. So the red graph should be above the X axis from zero point five, and we see that's exactly what's happening. So these conclusions suggest that our answer's correct

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