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JH
Numerade Educator

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Problem 52 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^5 x \cos^3 x dx $

Answer

$$\frac{1}{6} \sin ^{6} x-\frac{1}{8} \sin ^{8} x+C$$

Discussion

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Video Transcript

here will evaluate the indefinite integral of sciences of his power times. Cho sang cute. Then we'LL go out and graph the Interbrand circled in red as well as the whole anti derivatives, and will use the graft to check that our answer is reasonable. So first, let's just go ahead and evaluate the integral. So it's quite and rewrite this as signs of the Fifth Power and let's write Coastline Cubed is co sign squared Times Co sign, and then we could use the faggot identity to rewrite co sign squared. So let's go ahead and do that. Signed to the fifth so co sign squared is one minus sense where Times CO sign. So this suggests that we go ahead and use the U substitution. Let's go ahead and take you to be signed then to you as co sign next to the X. So we have integral you to the fifth one minus you squared Do you so used to the fifth minus use of the seventh and we'LL integrate this So Ian, you know the six over six minus you to the eighth power over eight Plus he And since we're taking C to be zero from the instructions we could go out and trump to see. So now we backs ups too, using our use of with signs of the six power over six, minus signs of eight power over eight. And there's our answer. So scored in color. Coordinate this with blue so the Inter Grand was in red. The anti derivative is in blue, so this is so the integral of the red is blue. That means that the derivative of the blue is red. So let's observe this on the next page. So we have our blue here, science of the six power over six, minus signs to the eight power over eight and this should be the original Intolerant, which was signed to the fifth time's Cho Sang Cube. So let's check if this is regional reasonable by going to the graph so again into grand in red, anti derivative and blue. So here the colors there consistent with how we label them. And it's a grand in red anti derivative in blue. So the red graph is supposed to be the derivative of the blue graph, so let's check every time. If the blue graph has a horizontal tension. If the bread graph zero. So here at zero, we see the blue graph has a horizontal tangent. Did the rivet of zero. And that's exactly what the red graph is at that point over here at around one point five, we see that we have a horizontal tangent on the blue. That means that the red graph should be zero. It is, and same argument holds that negative one point five. The horizontal tangent at the Blue Line means that the derivative is zero and the red graph zero. And maybe we could do a little more arguing here to commit ourselves. For example, when x zero all the way until X is about one point five, we see that the blue graph is strictly increasing. That means that the derivative is bigger than or equal to zero. And that's what exactly is happening with the red graph. We see that the red graph is above the X axis from zero to one point five. So this suggests that there are answers, Grant