2–74. Integration techniques Use the methods introduced in Sec...

Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

57. ∫ (from 0 to √3/2) 4/(9 + 4x²) dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are given the definite integral from 0 to 1 of 5 divided by 13 plus 5 X squared. D X Now, in order to approach this problem, we want to simplify this integral in such a way that we can apply the tangent inverse identity. The first thing we can do is we can reorder the terms in the denominator and factor out of 5. If we were to do those two steps, we get the integral from 0 to 1, of 5 divided by 5 multiplied by X2, plus 13 divided by 5. DX. And this will further simplify to leave us with the integral from 0 to 1 of 1 divided by X2 plus 13 divided by 5DX. Now, we can simplify this integral by using the tangent inverse identity. If we have an integral of the form X2 plus A2, this is going to simplify to be 1 divided by a tangent inverse of X divided by A. In this case here, our A term for this integral is going to be a squared is equal to 13 divided by 5, and this term is going to give us an a term of the square root. Of 13 divided by 5. And this is going to simplify to be the square root of 65 divided by 5. So, by using this identity, we can rewrite the current integral to be 1 divided by the square root of 65 divided by 5, multiplied by the tangent inverse of X divided by the square root of 65 divided by 5, and this is going to be evaluated from 0 to 11. Now, if we were to further simplify the given identity, we can rewrite this to be 5 divided by the square root of 65, multiplied by the tangent inverse of 5 X divided by the square root of 65, evaluated from 0 to 1. By plugging in the values of the boundaries, we have 5 divided by the square root of 65, multiplied by tangent inverse of 5 multiplied by 1, which is 5 square root of 65. Now, if we were to plug in 0 for X, we will get tangent inverse of 0, which is going to give us 0. So, plugging in 0 into the equation gives us a value of 0. So, all we need to do now is just simplify the current tangent inverse function, and that is going to give us a final solution of the square root of 65 divided by 13, multiplied by the tangent inverse function of the square root 65 divided by 13, and this is going to be the final solution to the given definite integral. And with that being said, the overall solution is going to be C. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.57. ∫ (from 0 to √3/2) 4/(9 + 4x²) dx

Watch next

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
57. ∫ (from 0 to √3/2) 4/(9 + 4x²) dx