Evaluate the integrals in Exercises 37–46.

∫(...

Question

Evaluate the integrals in Exercises 37–46.

∫(2θ + 1 + 2 cos (2θ + 1))dθ

Accepted Answer

Welcome back everyone. In this problem, we want to evaluate the integral of 3 x + 2 + 6 multiplied by the sin of 3 x + 2, all with respect to x. Now we can evaluate this integral by using substitution. OK. Now, by substitution, OK, let's set you to be equal to 3 x + 2, the argument of our sine function. Well, in that case, that means if we differentiate here, the derivative of u with respect to x will be equal to 3. So duu is going to be equal to or sorry, rather dx is going to be equal to a third of duu. So now, in this case, we've been able to find or write UX in terms of you, OK, and we've also been able to write DX in terms of DU. So if we apply that principle to help us substitute, then by replacement. This would mean that the integral. Of 3 x + 2 + 6 multiplied by the sine of 3 x + 2 with respect to x. Is going to be equal to the integral of u. Rememberu replaces 3 x plus 2. Plus 6 multiplied by the sin of you, OK, multiplied by 1/3 DU. No, here, if we were to factor out to our constant. Then we'll end up with a third of the integral of you plus 6 sign you with respect to you. And now if we apply the linearity of integrals, then this is going to be equal to 1/3 of the integral of UDU. Plus the integral of 6 sin u d u. Now we can go ahead and integrate. If we start with the integral of UDU, OK, it's going to be equal to half of U2 using the antiderivative rules. And well, plus some constant C1 and the integral of 6 sin UDU is going to be equal to -6 u, OK, plus some integral C2. No, if we were to combine our results here, then we'll end up with 1/3 multiplied by 1. Well, we combine and we multiply throughout, we'll end up with 1/6 rather of u square minus 6 divided by 3 or 2 costs you, and we can combine our constants to rewrite it as c. Now remember, we know that U equals 3 x plus 2. So if we put that into our integral then. We'll have 1/6 of 3 x + 22 minus 2 multiplied by the cosine of 3 x + 2 + C. Thus, this is going to be the integral of 3 x + 2 + 6 sin of 3 x + 2 with respect to x. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 37–46.∫(2θ + 1 + 2 cos (2θ + 1))dθ

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Evaluate the integrals in Exercises 37–46.

∫(2θ + 1 + 2 cos (2θ + 1))dθ