the integral from -1 to 2 of (cost +2tsint) dt

The indefinite integral first:

INT.[cost +2t*sint]dt

= sint +INT.[2t*sint]dt

Do the second integral by parts: INT.[u]dv = uv -INT.[v]du

Let u = 2t, so, du = 2dt

And dv = sint dt, so, v = -cost

= sint +2t*(-cost) -INT.[-cost](2dt)

= sint -2tcost +2sint +C

= 3sint -2cost +C -------------(i) <--------edit: should be -2tcost.

Then the definite integral:

INT.(-1 --> 2)[cost +2t*sint]dt

= [3sint -2cost] | (-1 --> 2)

= [3sin(2) -2cos(2)] -[3sin(-1) -2cos(-1)]

= [3sin(2) -2cos(2)] -[3sin(-1) -2cos(-1)]

= 7.1652 -----------------------------------answer.

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Edit: Sorry, but I forgot about the t in -2tcost.

So let me correct the computations,

Then the definite integral:

INT.(-1 --> 2)[cost +2t*sint]dt

= [3sint -2tcost] | (-1 --> 2)

= [3sin(2) -2*2*cos(2)] -[3sin(-1) -2(-1)cos(-1)]

= [3sin(2) -4cos(2)] -[3sin(-1) +2cos(-1)]

= 5.836288 -----------------------------------answer.