In Exercises 67–73, use integration by parts to establish the ...

Question

In Exercises 67–73, use integration by parts to establish the reduction formula.

∫ x^n sin(x) dx = -x^n cos(x) + n ∫ x^(n-1) cos(x) dx

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Welcome back everyone. Evaluate the integral x to the power of n, ln of x, d x4 n not equal to -1, and x greater than 0 by integration by parts and give the indefinite integral in closed form. For this problem we're going to use integration by parts. Remember that the integral of UDV is equal to uv minus integral of VDU. We're going to set the u equals ln of x, and dV is going to be the remaining part which is x to the power of n. D X Now let's identify DU by taking the derivative DU divided by dx. That's the derivative of L of X, which is 1 divided by x. We can multiply both sides by dx to show that DU is equal to dx divided by x. From here we also need v, which is the integral of dv, and as the integral of x to the power of n d x. Here we can use the power rule because n is not equal to -1, so we get x to the power of n + 1. Divided by n plus 1, no constant of integration is needed for now, right? And we can now show that the integral of x, the power of n ln of x, d x is going to be equal to u multiplied by v. So that's X the power of n + 1 divided by n + 1. That's v multiplied by u, so multiplied by ln of x, and we're subtracting the integral of VDU. So we can take x the power of n + 1 divided by n + 1. That's V and DU is equal to Dx divided by x. So we're multiplying by dx. Divided by x. We can now simplify, specifically, we can factor out the constant, right? And we're going to get. X to the power of n + 1. Divided by n + 1. L n of x, that's our first part. For the second part, we can take out 1 divided by n plus 1, which is our constant. And we can integrate x to the power of n + 1 minus 1 because we are dividing by x, right? So that's the integral of x to the power of n d x. And we already know what it is, right, so we can just substitute the result to get x, the power of n + 1. Divided by n + 1. Alan of x minus. 1 divided by n plus 1 multiplied by the integral which is X the power of n + 1 divided by n + 1. And we're going to add a constant of integrationc because that's an indefinite integral, right? And now let's go ahead and multiply to get x the power of n + 1 divided by n + 1 multiplied by ln of x. Minus X to the power of m + 1 divided by M + 1 squared Plus a constant of integration c and this is how we arrive at our final answer for this problem. Thank you for watching.

In Exercises 67–73, use integration by parts to establish the reduction formula.∫ x^n sin(x) dx = -x^n cos(x) + n ∫ x^(n-1) cos(x) dx

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In Exercises 67–73, use integration by parts to establish the reduction formula.
∫ x^n sin(x) dx = -x^n cos(x) + n ∫ x^(n-1) cos(x) dx