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Evaluate the integral.

$\int_{0}^{a} \frac{d x}{\left(a^{2}+x^{2}\right)^{3 / 2}}, \quad a>0$

$\frac{1}{\sqrt{2} a^{2}}$

Integration Techniques

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we have the definite integral from zero to a of one over a squared plus X squared to the three half power where a is a positive number. So by looking at our denominator, we see a square plus x square, so we should take the tricks up to be X equals a tan theta. Therefore, D X's A C can square data the data. So observe here we have a definite rule, so we have two options on how to proceed. The first option is we can find the new limits of integration in terms of the variable data. If we do that, well, we won't have to draw the triangle because we won't need to go back to the variable X. The second option is to not find the new limits of integration. Then you would have to evaluate the anti derivative draw the triangle back to the Variable X, and then you could plug in these X values into the anti derivative. So let's see if we can take the first round so we don't have to draw the triangle here, so we need to find a new limits. So the lower limit and also before I go on. We have to make an observation here, and it comes from the tricks up. If you want to use this method, you have to be cautious about the the domain for data. So here, when we do this type of tricks of the T in substitution, your data lies between negative proper too, and powerful, too. So the lower limit. So before we had X equals zero. So plugging in X equals zero into this equation. Up here we have zero equals eight and data since a is positive number, that was the assumption. This means that tan has to be zero and the only time that tangent zero from negative pi over 22 pi over two is when data equals zero. So this is our new lower limit in terms of data. Similarly, for the upper limit, this was X equals a so plugging this into our tricks of equation. We have a equals a tan data, which means tan data is one and the only time that happens in this interval. Negative pi over 22 pi over two is when data equals power before mhm. So let's plug all this in. We have the integral from zero to power before DX becomes a C can squared data, data data. And in the denominator, we have a squared plus X squared, which is a squared 10 square data. And this is also the three house. Yeah, So at this point, let's take a look at this denominator here. And let's rewrite this before we before we go on. So here we can we can pull out of a squared, then we have one plus hand square data. So we have a squared Seacon square data to the three halfs and that will become a cubed times seeking cube data. So, yeah, now we could simplify. Let's cancel. So we we can cross off this a with one of those. So we have a square left over and we could cancel out those to see cans and then re elected with one second on the bottom. So we have one over a square, integral zero to pi before of one over. C can't data. Now. I'm running out of room here, so let me go to the next page One over a square, integral zero power before and then one oversee can't. By definition, it's called Santa. Yeah. So now we have we can evaluate this anti derivative. We know that science data and we have our endpoints. So now we just use the union circle to evaluate these. Sign of pyro for is square root 2/2 and then sign of 00 So there's our final answer.