79–82. {Use of Tech} Double table look-up The following integr...

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79–82. {Use of Tech} Double table look-up The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system.
82. ∫ (sin⁻¹(ax)) / x² dx, a > 0

82. ∫ (sin⁻¹(ax)) / x² dx, a > 0

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says to evaluate the integral and we're given the integral of the inverse cosine of the quantity of BY and quantity divided by Y2 Dy where b is greater than 0. So to evaluate this interval, we first want to make a substitution. So we're going to set U equal to the inverse cosine of the quantity of BY in quantity. And if we solve this equation for Y, the first thing we'll need to do is to take the cosine of both sides of the equation. We'll have cosine of U equal to the Y and divide both sides by B, we get that Y is equal to cosine of U divided by B. And we can use this expression to show that DY is going to be equal to minus sign of U divided by BDU. So making the substitution, we see that are integral. Of the inverse cosine. Of the quantity of BY in quantity, divided by Y 2 D Y is going to be equal to the integral. Of you divided by the quantity of cosine of u divided by b and quantity squared multiplied by minus sine of u divided by B multiplied by D U. And so simplify this expression, this is going to be equal to the integral of minus B multiplied by U, multiplied by the sign of u, divided by the quantity of cosine squared of UDU, and we can use our trigger identities to rewrite this integran. So this is going to be equal to, and we're actually going to pull the constant out in front of the integral, so we'll have minus B multiplied by the interval. Of you, multiplied by the tangent of you, multiplied by the sequent of U D U. Now for this integral we can actually integrate by parts. And so we'll set F equal to you, and that means that DG is going to be equal to the tangent of U multiplied by the sequant of U DU, and that means that DF is going to be equal to DU, and that means that G is going to be equal to the sequent of U. So integrating by parts, we see that our integral here is going to be equal to minus B multiplied by the quantity, and here we integrate by parts recall that we'll have F multiplied by G, so that's going to be U multiplied by the sequent of U, then we'll have minus the integral of GDF. So that's going to be sequent of U. Do you In quantity So now we just need to integrate that second interval. And so that means that this is going to be equal to. Minus B Multiplied by the quantity of you, multiplied by sequent of you. Minus, and here we call that when we integrate the second view with respect to you, that we get the natural log of the absolute value of the quantity of the tangent of you, plus the sequent of you. In quantity Plus C. So now we need to write everything in terms of Y, since our final answer should be in terms of Y. So since U is equal to the inverse cosine of the quantity of By, if we look at sequent of you, that is going to be equal to the sequent. Of the inverse cosine of the quantity of BY. In quantity, and that is going to be equal to 1 divided by the quantity of By. And the other term that we need to look at is going to be the tangent of you, and this is going to be equal to. And here we're going to write the tangent of you as the sine of view divided by the cosine of view. So that means that tangent of view is going to be equal to the sign of the inverse cosine. Of quantity of By in quantity divided by the cosine of the inverse cosine. Of the quantity of By in quantity. Which is going to be equal to. The square root Of the quantity of 1 minus the quantity of BY in quantity squared in quantity divided by the quantity BY in quantity. So now, making these substitutions and back to our expression that we have calculated, we see that the interval. Of the inverse cosine of the quantity of BY in quantity divided by Y 2 D Y is going to be equal to minus B multiplied by the quantity of the inverse cosine of quantity of BY in quantity. Multiplied by One divided by the quantity of BY in quantity. Minus The natural log of the absolute value of the quantity of the square root of the quantity of 1 minus the quantity of BY in quantity squared in quantity divided by the quantity of BY in quantity. Plus 1 divided by the quantity of BY in quantity, plus C. And so now we just need to simplify this expression. So distributing the minus B through, we see that this is going to be equal to B multiplied by the natural log of the absolute value of the quantity of the square roots of the quantity of 1 minus the quantity of BY in quantity squared in quantity plus 1 in quantity divided by the quantity of BY in quantity. Minus the inverse cosine. Of the quantity of BY in quantity, divided by the quantity of Y in quantity plus C. So this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

79–82. {Use of Tech} Double table look-up The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system.82. ∫ (sin⁻¹(ax)) / x² dx, a > 0

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79–82. {Use of Tech} Double table look-up The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system.
82. ∫ (sin⁻¹(ax)) / x² dx, a > 0