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JH

# Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle.$\int \frac{ x^3 }{ \sqrt{x^2 + 4}\ } dx$$x = 2 \tan \theta$

## $\sqrt{x^{2}+4}\left(\frac{x^{2}-8}{3}\right)+C$

#### Topics

Integration Techniques

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

let's evaluate the integral using the indicated trick substitution. So here we have a integral of X cubed over this women of X squared plus four in our tricks of is X equals to tan data. This was provided. So let's find the X the exes too. No seek and square data data. So using this tricks up, we have the following the original integral equals. So now first thing is we could replace X cubed with two tan cubed. So we have two cubed, which is eight tan cute and then DX is too sequence where there aren't any data. Hm. Now, in the denominator, we have X square plus four, So explore it becomes fourth and data square. Let's go ahead. And before we go on, let's go to the side and simplified the radical so we can pull out the four outside the radical that becomes a too. Then we have square root of transfer data plus one which is two square root of seeking square data which is to seek and data also we can cancel off this to see can't with this too, and one of the sequence in the numerator so we can go ahead and cross this off. We can take off this too. And we could lose one of those. He can't. So we have eight integral tan Kim Saito seek and data. So I have a trick in a metric and rule. Let's go ahead and proceed because we have seeking of to the first power. Let's rewrite this by pulling out one of the engines. Now we can use the U substitution after. Maybe the first thing we should do is rewrite Tangent Square so we can rewrite Tangent squared is six square data minus one and then here we have the derivative of C can't. So this is why I will use a U substitution. So here we should take you to be sick and data So that do is seek and data tan data and are after these of our general becomes eight. You squared minus one. Do you? So I'm running out of room here. Let's go to the next page. No. So picking up where I left off take anti derivative minus eight times you plus our constant of integration and then we use are you sub which was toe back substitute from you back into data So we have eight over three c Can't you minus eighty Cantero So if the fund original problem was given in terms of X, we'd like to go back to X. So to do so, we'LL have to use our u substitution, which was X equals to tan data which implies tan data is X over too. So let's make a right triangle out of this. So there's data so tan data is opposite over adjacent. So let's take X into and by the category in the room. If we call this high partners age, we know that a square is X squared plus two squared So h is the square root of X squared plus four So now we have all the sides of the triangle we can evaluate, seek it and seek and Cube. So speaking of data is high partners over adjacent silly have X squared plus four over to So they're seek it. But then we're cubing this so to the third power minus eight h over too. So here we could simplify this a bit a delivery on its numerator. We have X square plus four to the three have power and then two to the third Power is eight sweet across those off and then ate over fours to just excuse me eight over to us for so we have X squared plus four to the three half power over three minus four. And then we have a X squared plus four plus e and we can't finish here. But at this point, we could factor out, sees me hit this should have been a square root. At this point, we can go ahead and factor out the square root of X squared plus four. Let's go to the next patient. Do that. So we're left over with X squared plus four over three minus four and then we can just go ahead and simplify this. So we'LL have a X squared minus c and there's a final answer.

JH

#### Topics

Integration Techniques

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp