Evaluate the integrals in Exercises 1–24 using integration by ...

Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ θ cos(πθ) dθ

Accepted Answer

Welcome back everyone. Compute integralt cosine 3 TDT using integration by parts. For this problem, let's begin with the formula of integration by parts. Integral of UDV is equal to UV minus integral of VDU. What we're going to do is set u equals t. And the remaining part, which is dV is going to be cosine of 3 t dt. What we want to do is identify DU, which is dt because u equals t, right, and V equals integral of dV. So that's integral of cosine 3 c dt, which is going to be equal to 1/3 sin. 3 C. We're not adding any constants of integration, right? So let's apply the formula now. The integral oft multiplied by cosine of 3 t. The t is going to be u multiplied by v. So that's t multiplied by 1/3. We can write it as t divided by 3 multiplied by sin of 3 t minus integral of VDU. And we know that V is equal to 1/3 sin 3T, so we can write 1/3 integral sin 3t, and now DU is DC. So now we want to integrate sin. Let's go ahead and do that. We're going to rewrite the first term, t divided by 3, sin 3c minus 1/3 multiplied by. Now the integral of sine is negative cosine. We're going to write negative cosine of 3 c, and according to the reverse chain rule, we're dividing by the slope of the linear term. So we're dividing by 3, and we're going to add a constant of integration c because that's an indefinite integral. Now let's go ahead and simplify C divided by 3. Sign 3 T. Plus 19. Cosine 3 t plc.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫ θ cos(πθ) dθ

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Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ θ cos(πθ) dθ