79. Tabular integration extended Refer to Exercise 77.
a...

Question

79. Tabular integration extended Refer to Exercise 77.

a. The following table shows the method of tabular integration applied to

∫ eˣ cos x dx.

Table showing functions and derivatives of f and integrals of g with arrows illustrating tabular integration steps.

Use the table to express ∫ eˣ cos x dx in terms of the sum of functions and an indefinite integral.

b. Solve the equation in part (a) for ∫ eʳ cos z dz.

c. Evaluate ∫ e⁻ᶻ sin 3z dz by applying the idea from parts (a) and (b).

Accepted Answer

Welcome back, everyone. In this problem, we want to use tabular integration to evaluate the integral of E to the X cost of 3 X with respect to X. Now, we're going to use tabular integration as we need to perform integration by parts multiple times on the same integra. That eliminates the need to constantly rewrite the integration by parts formula and manage multiple negative signs. So, by simply filling in two columns, one for derivatives and one for integrals, then we can multiply diagonally to find the. Pollution. That reduces the complexity and the chance of making a sign or a constant error. So, like we would normally do in integration by parts, first, let's use lipid to choose FFX and GFX. Because we know what it basically tells us is that we will set as FFX whichever one of these types of functions come first, whether it's logarithmic, inverse trigonometric, polynomial, exponential, or trigonometric. Now if we look at our product here, we have E to the X. That's an exponential function. The cosine of 3 X is a trigonometric function. Thus, E X would come first in our order. So we're going to choose F of X as e to the X, OK, which makes sense because This function E to the X, the derivative and in its integral is constant. And that means then we're going to set Geofix. To be equal to the cosine of 3 X or trigonometric function. Now let's make our two column table. So let me just make a little table here, OK, of FF X being E equal to E to the X and G of X being equal to the cosine of 3 X. Now, basically, what we're gonna do is that we are going to differentiate and integrate until we end up with the original term that we started with for G of X, OK? Until we get a product that's a constant multiple of this original integral. So in our first column, if we differentiate F of X, we get the derivative of F of X to be equal to E X, and if we let PX be equal to the integral of G of X, then it's the integral of the cosine of 3 X, which is a third of the sine of 3 X. We don't have or cosign a 3x term in this integral, so we'll differentiate and integrate again. So now we go to our second derivative of FF X, which would be either the X. And if we let QFX be the integral of PF X. Then it's the integral of 1/3 of the sine of 3 X which would be equal to negative 19 of the cosine of 3 X. So now, notice that in this 3rd row, we now have a constant multiple of the original integral of the cosine of 3 X, OK? So we can stop at our 3rd row. No, we're going to multiply our inches diagonally and add alternating signs. So we're gonna multiply F of X by P X, and that's going to be positive. Next, we're going to multiply the derivative of FF X by Q X, which is gonna be negative. And then, since there are no more rows, we're going to multiply QF X by the, the second derivative of FFX and we're going to make that positive, OK? So let me just move our plus sign. Aligned with our arrows. So no, what we've done is is to perform diagonal multiplication for all the rows leading up to the repeating function. So this product is not a term in the sum, but rather the integra of the remaining integral. So we're going to place this integral on one side of an equation and the diagonal products on the other, and then solve the resulting algebraic equation to find the value of the original integral. Let's actually do it just to show what that means. So no Means we're gonna take our integra. So the integral of E to the X multiplied by the cosine of 3 X with respect to x. And it's going to be equal to our first product. A third of E to the X multiplied by the sine of 3 X, OK? Plus our second product that is gonna be negative E to the X multiplied by 1/9 of the cosine of 3 X, which is gonna be plus 9 of E to the X multiplied by a cosine of 3 X. OK. Minus or integral, OK. Of 1/9 of e to the X multiplied by the cosine of 3 X with respect to X or remaining integral. Now, this is good because if you notice here, we can group our integrals together. So that means we'll have the integral of E to the X multiplied by the cosine of 3 X with respect to X plus 1/9 of the integral of E to the X. OK. Let's just put a little gap here. E to the x multiplied by the cosine of 3 X with respect to x and on our right hand side we are left with a third of e to the X multiplied by the sign of 3 X plus 1 9in of EX multiplied by the cosine of 3 X. Now for our two terms here, if we're adding 1 to 19th of this term, it's going to be 109, OK, of the integral of E to the X plus 3 X with respect to X. And on our right hand side, if we group or if we factor out 1/9 of E to the X, then we're left with 33 X plus plus 3 X. Now if we multiply both sides of our equation, OK, by 9/10, the reciprocal of 109, 10/9s will cancel on the left hand side, which just leaves us with the integral of e to the X multiplied by the cosine of 3 X with respect to X. Being equal So over here are 9s cancel, so that's equal to 10th of E to the X multiplied by 33 X plus the cosine of 3 X. Thus, this is our integral of E to the X multiplied by the cosine of 3 X with respect to X. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Tabular Integration Method

Integration by Parts

Solving Integral Equations Involving the Same Integral

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