9–40. Integration by parts Evaluate the following integrals us...

Question

9–40. Integration by parts Evaluate the following integrals using integration by parts.

20. ∫ sin⁻¹(x) dx

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the following integral using integration by parts the integral of inverse cosine of XDX. Fantastic. So it appears for this particular problem we're ultimately trying to evaluate this particular integral that is given to us by the problem itself using integration by parts. So now that we know that we're ultimately trying to evaluate this integral by using integration by parts, our first step that we need to take in order to solve this problem is we need to recall and use lippi. Which is an acronym for LIPET, and this is to help us choose the values of U and DV. So, as we should recall, Lipit stands for L I PET, which means that L stands for logamithic function, so like LNF X. I stands for inverse trigonometric functions such as Inverse tangent of X. P stands for polynomial, such as X to the power 2 or Y, etc. E stands for exponential, such as E to the power of X, and T stands for trigontric, such as sine of X or cosine of X, etc. So once again, we're trying to use this acronym Lipi to help us choose our values of U and DV. So we need to be able to recognize that as we're told by the prom itself, so we have this inverse cosine, and we're told that inverse cosine of X is given to us, and we have the integral of inverse cosine of XDX, and as we should be able to recognize. Inverse cosine of X is an inverse trigometric function, so we need to select U is equal to inverse cosine of X. And we need to recall a note that DX is just 1, so we need to select that DV is equal to DX. So with that in mind, our second step now is we need to differentiate you and integrate DV. So U is equal to inverse cosine of X, which will mean that DU is equal to -1 divided by, so we need to say that we need to note that DU is going to be equal to -1 divided by the square root of 1 minus X to the power of 2. DX. We also need to note that DV is equal to DX, so that will mean that V is equal to X. Our third step now is we need to recall and apply the integration by parts formula, which, as we should recall, that the interval of UDV is equal to U multiplied by V minus the interval of VDU. Fantastic. So, moving right along here, that will mean that we need to use at this point, we need to use substitution. So we need to take the integral of inverse cosine of X D X is equal to X multiplied by inverse cosine of X minus the integral of X multiplied by parentheses -1 divided by the square root of 1. Minus X2. D X, and we need to note that this will be equal to x multiplied by inverse cosine of X plus the integral of X divided by the square root of 1 minus X 2 DX. Fantastic. So moving right along here. Our 4th step now is we need to evaluate the integral X divided by the square root of 1 minus X2. Fantastic DX. So we're trying to evaluate this particular integral, and we need to use substitution again. So let's note that we will let W equal 1 minus X to the power 2. So we could say that DW is equal to -2 XDX, which we could say is -12, DW is equal to XDX. So with that in mind, we now need to rewrite our integral. So we need to take the integral. So once again we're rewriting our integral, so we need to take the integral of X divided by the square root of 1 minus X2 DX is equal to the integral of x divided by the square root of W. D X, which is equal to the integral of negative 12 DW divided by the square root of W, which is equal to negative 1/2 multiplied by the integral of W to the power of negative 1/2. DW. Fantastic. So, moving right along here. We now need to integrate, and when we integrate, we will find that it's equal to -12 multiplied by 2 to the, so we need to take. And our integral and we're gonna integrate, and when we integrate, we will find that it's equal to -12 multiplied by 2 to the power of 12 plus C, which is equal to negative square rooted W plus C, which is equal to negative square root of 1 minus X to the power of 2 plus C. So now, we need to put it all together, and when we do, we will find that the integral of inverse cosine of XDX is equal to X multiplied by inverse cosine of X minus the square root of 1 minus X to the power 2 plus C, and that is our final answer. Hooray, we did it. We saw for our final answer. So Looking at our problem itself, we can summarize that our final answer, without a doubt, once again, will be X. So once again, X multiplied by inverse cosine of X minus the square root of 1 minus x 2 plus C. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

9–40. Integration by parts Evaluate the following integrals using integration by parts.
20. ∫ sin⁻¹(x) dx

Key Concepts

Integration by Parts

Inverse Trigonometric Functions

Differentiation and Integration of Algebraic and Transcendental Functions

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