Evaluate the indefinite integrals using Substitution. (use C...

Evaluate the indefinite integrals using Substitution. (use C for the co a) \int 3x^2(x^3 - 3)^3 dx = b) \int (2x - 5)(x^2 - 5x - 9)^6 dx = c) \int x(x^2 + 1)^4 dx = d) \int -(8x + 10)(- (2x^2 + 5x + 8))^4 dx =

Transcript

00:01 So in this problem we are given three parts let's start up with part a part a we want to evaluate the integral of cosine to the fourth power of three x so i'm going to write out a cosine of three x all of these to the power of four times sine of three x dx notice that i'm going to replace um this in terms of u let u be equal to cosine of three x the u is going to equal to the little of cosine is negative sine of 3x times the root of the inside is just 3 times dx so therefore negative 1 3rd times the u is equal to sign of 3x d x so this integral is now going to equal to the integral of so let's write out negative 1 3 3rd first the integral of u to the power of 4 the u so this is just going to be negative 1 3 3 3 of um 1 5th of u to the power to 5 plus some arbitrary constant c which is going to be equal to negative 1 over 15th times cosine to the fifth power of 3x plus some arbitrary constant c this is pretty much the answer for part a let us move to part b so for part b we want to evaluate the integral from 0 to 2 of x times e to the power of 2x the x the x the x so here we're going to do something similar so let you be equal to 2x sorry in this case we're going to do something very similar to what we did before except that we are going to be using integration with parts so using integration by parts we are going to let you be equal to x the u is going to equal to 1dx dv is going to equal to e to the power of 2x d x and v is going to equal to 1 half e to the power of 2x so i draw my 7th over there so this integral is now going to equal to u times v so that's x times 1 half e to the power of 2x evaluating this from 0 to 2 minus the integral of v to u that is 1 half integral of e to the power of 2x d x d x where the integral is from 0 to 2.

02:40 So we're going to have, let's plug in 2 first.

02:42 We're going to have 2 over 2, which is 1 times e to the power of 4 minus 0.

02:48 When we plug in 0, we just simply get 0, minus 1 1 1⁄2 of e to the power of 2x, evaluating this from 0 to 2...

Question

Evaluate the indefinite integrals using Substitution. (use C for the co a) \int 3x^2(x^3 - 3)^3 dx = b) \int (2x - 5)(x^2 - 5x - 9)^6 dx = c) \int x(x^2 + 1)^4 dx = d) \int -(8x + 10)(- (2x^2 + 5x + 8))^4 dx =

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