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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 x \sin^2 (x^2) dx $

$\frac{1}{4}\left(x^{2}-\sin x^{2} \cos x^{2}\right)$

Integration Techniques

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for this problem will evaluate the indefinite integral. And then we'LL check that our answers reasonable by graphing both the immigrant and the integral. So first, let's evaluate the integral here it probably be best toe start off with a U substitution. Let's take you to be x square then do you? Is two ex tx so that do you over too this x t x So using our use up, we have one half integral sine squared of you, do you? And then we could use our trick in a metric identities to rewrite. This is one minus co sign up to you all over too. So again, that's coming from Trig double angle formula so we can pull out this too at the denominator. So we have a one over four and then the integral of one with respect to you is just you minus And then the anti derivative of co sign becomes sign of to you over too. And here for this integral. If this too was bothering you, you can go ahead and do another U substitution here You can take w to be to you to arrive at this anti their evidence. Then the next thing we could do is come back to our U sub and replace you with X squared. So we have one fourth X squared minus sign of to X squared over to and also because we're taking C to be zero. There's no need to right see anymore. So it's good and cross that off and then you could basically stop here if you like. I would like to simplify this a little more, but using the double angle formula for sign, not necessary. But it just simplifies things. So we have one over four X squared minus sine X squared coastline explored. And the reason I didn't write the Tuesday is because the two that we would get from this double angle formula cancels with the two that was in the denominator. So Scott and write our final answer one fourth X squared, minus sign of X squared coastline of X squared. So let's go ahead and let's we're gonna graft are our inside grand. So it's color that and read the ex sine squared of X squared, and we'LL also graft the anti derivative, which was our final answer Over here. Let's scrap this one in blue So now one observation here is since the anti derivative of the red equals blue That means that the derivative of the blue equals the red So we can graft thes And then we could see if the observation is true. So ideas here let's summarize that. So we know to expect on the graph so mean one more page over. So our observation something to look out for is that Thanks Sign Square X squared is the derivative. So again, I'm using fundamental dilemma Kokkalis here. So this is our blue blue graph. There's a right Graff. So go to the graph and see if this is true. So here we're color coordinated. So red is still the intolerant and blues, the anti derivatives. So we concluded that the red graph is the derivative of the blue graph. So let's see if that's true here. So, looking at the graph, we see that, for example, over here, where the cursor is that we see that the blue graph has a horizontal tangent. That means that the derivative of zero and that's exactly where the photograph is. That is that zero same thing over here at around two point five we see that the blue graph has another horizontal tangent. The derivative is zero, and that's exactly where the road graph is, And we can also do that on the negative side. Over here. Betweennegative one point five and negative, too. We see another horizontal tangent on the blue, so it's derivative is zero, and that's where the red is, and we could keep doing that. And that also happens here at the origin Horizontal tangent. On the blue graph read a graph of zero. I also noticed that for on the right hand quadrant, so when X is bigger than zero that the blue graph is always increasing. That means that the derivative is bigger than or equal to zero. And that's what's happening with the red graft. The paragraph is strictly above the X axis, and similarly, on the left, we see that the blue graph is always decreasing, so that means that the derivative is always less than or equal to zero. And that's exactly where the photograph lies. So this suggests that our answer was correct.