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Numerade Educator



Problem 1 Easy Difficulty

Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle.

$ \displaystyle \int \frac{dx}{x^2 \sqrt{4 - x^2}} $ $ x = 2 \sin \theta $


$-\frac{\sqrt{4-x^{2}}}{4 x}+C$


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Video Transcript

here will evaluate the inaugural using that indicated triggered a metric substitution. We have the integral of one over X where times square root of four minus X squared. They already have provided us with different. So So from district sub, we have one. The ex is too cosign Data, data, data. So the integral let's rewrite this so the ex becomes to co sign data data and in the denominator, we have X squared. So that's it for sign square and then we have square root of four minus for science flor. So here, let's go ahead and simplify this radical we have so we could pull out the four here that becomes too one minus sign square that's co signed square. And this is equal to two times co signed data, so that takes care of the radical. So it's going and cancel this to co sign data with the two coasts and data on the numerator. Then we have one over four in a girl data over science cleared, which is one fourth in rural. Of course, Higgins, where what and we know that this is integral of cosi cans is squared is negative attention. So we have negative co attention data over four. Plus he. And now, at this point, we'd liketo use the right triangle corresponding to the transom. Let's come over here and do this. So this original substitution means that signed data is extrovert too. So let's draw any right triangle that has this property. So if we do not, they stated down here in the bottom, right? Signed data is X divided by two. And then we can use the Pythagorean Darryl. So find this bottom side. Let's call it tea. We know T square plus explorer is too square. So that means he is the square root of four minus x square. So now, since we have all the signs of this red triangle, we can evaluate co tension of data. So we have negative, So co tension is adjacent over opposite. So we have tea over X so radical four minus x square, which was t over X And then also we have this for in the bottom and then plus e. And there was a final answer