Evaluate the definite integral. 0∫ π/ 2 sin ( 2 t ) d t

Question

Evaluate the definite integral.0

lower limit; 0

upper limit: pi/2

I tried taking the antiderivative and got -cos(2t^2)/2 which simplifies to -cos(t^2) then plugged int the intervals to get -cos(pi/2)^2-cos(0)^2 =- 1.99925 which is wrong.

what did i do wrong? was it a math error??

Answer 1

Let u = 2t, then du = 2dt. When t = 0, u = 0; when t = π/2, u = π.

∫₀^π/2sin(2t)dt =1/2 ∫₀^πsin(u)du = 1/2 [-cos(u)|₀^π] = -1/2 [ cosπ - cos0] = -1/2 [ -1 - 1] = -1/2[-2] = 1

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