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Evaluate the integral.
$ \displaystyle \int \frac{dx}{[(ax)^2 - b^2]^{\frac{3}{2}}} $
$-\frac{x}{b^{2} \sqrt{(a x)^{2}-b^{2}}}+C$
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 3
Trigonometric Substitution
Integration Techniques
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here we have the integral of one over a X in parentheses, squared minus B squared all to the three halfs power. So looking in the denominator, we see this a X squared minus B squared. So that suggests our tricks up. Should be a X equals B c can data and then solving for X. So we divide by a then we differentiate, right? Yeah, to get RDX in terms of detailer And now, before we plug everything in let's just go ahead and take this original denominator that we have And let's go ahead and simplify that. So we have X squared minus B square. So the three halfs power So now, using our tricks of a X is equal to B c can. So this is d squared. C can square minus b square. Yeah, let's go ahead and fact throughout the B squared and then we'll go ahead and use one of the powers one of the exponents properties to rewrite. This is B squared three halves and then c can squared minus one is tangent squared. So we have Tangent square also to the three halfs. So then we have be cubed Time stand cute All right, so let's plug all this in. So the X that was B over a seek and data and data The data. We have those from our step over here and then in the denominator. We just simplified that. And that's B cubed tam. Cute. Uh huh. Simplify as much as we can. We see, we could take off one tangent. So let's take off one in the bottom. That's it to left over. And then here we could lose this, be on top, and then we replace this power down here with another two, right? Come, let's go and pull out the constants. So we have a one over a times B squared in front of the integral and then on top were left over with. See, Cantero did ERA, whereas in the bottom, we still have 10 square Now rewrite c can is one of our coastline, and here we can write 1/10 squared as co sign Square Oversight Square. Yeah. Yeah, And then we can go out and cancel one of the co signs. Cancel this one and then you slept with a one up top co science is the first power. Mhm. So we have co sign data sine squared theta D data. So let me go to the next page. I'm running out of room here. Yeah. So we had co sign on top Science Square on bottom and for integral of this forum, it's probably best to go out and easy use sub sticking you to be signed data. Then do us the numerator coastline data the data so we can write. This is one over a B squared, integral one over you square. So this is the one over Science Square, and then the d you gives me the coastline dictator Use the power law. The power rule to integrate this. Yeah, and then back substitute to obtain negative one over a B square. You assign data so I could replace this with the data. Okay, so we've integrated in terms of you. Act substituted from this use of up here to get you back in terms of data, but the original problem was posed in terms of X. So we have to draw the triangle involving data so that we can evaluate sign back in terms of X. So remember our tricks up. Mm. A X equals B C. can. Taito. That means seek and data is a X over B. So let's try to draw right triangle using this information so she can't. It's high partners over adjacent. So let's take this to be a X and let's take the adjacent to BB. If the missing side here on the right is on the left, the stage by Pythagorean theorem we have a squared plus B square equals X squared, and then we could solve that for age. So now we have all three sides of the triangle so we can evaluate any trade function so we'll definitely be able to evaluate the sign here. So this becomes negative one a B squared and then for sign. Well, let's go ahead and write this because signs in the bottom. So we still have signs over here Sinus H over a X so that that would be the radical. And now we can go out and simplify this a little bit so we could cancel those A's. And then we could go out and put this X back in the numerator. So we'll have negative X b squared radical, a X squared minus B square and then plus C mhm. And there's a final answer
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