Question

Evaluate the integral. π/2 0 3 sin2(t) cos(t) i + 3 sin(t) cos2(t) j + 6 sin(t) cos(t) k dt

Answer 1

To evaluate the given integral, we need to simplify and integrate each term separately by expanding the trigonometric expressions. The evaluated integral is (-3/4cos^2(t) + C1)i + (3/8sin^2(t) + C2)j + (-3/4cos(2t) + C3)k + C, where C1, C2, C3, and C are constants of integration.

Step-by-step explanation:

To evaluate the given integral, we need to simplify and integrate each term separately. Let's start by expanding the trigonometric expressions:

3sin^2(t)cos(t) = 3/2sin(2t)cos(t)

3sin(t)cos^2(t) = 3/2sin(t)cos(2t)

6sin(t)cos(t) = 3sin(2t)

Now, we can integrate term by term:

Integral of 3/2sin(2t)cos(t) dt = -3/4cos^2(t) + C1

Integral of 3/2sin(t)cos(2t) dt = 3/8sin^2(t) + C2

Integral of 3sin(2t) dt = -3/4cos(2t) + C3

Therefore, the evaluated integral is (-3/4cos^2(t) + C1)i + (3/8sin^2(t) + C2)j + (-3/4cos(2t) + C3)k + C, where C1, C2, C3, and C are constants of integration.