What is the integral from 0 to t of 2sin(t) ?

The integral from 0 to t of 2sin(t) is 2(-cos(t)+1)

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integral from 0 to t of 2sin(t)

\int_{0}^{t} 2 sin (t) d t

2 (- cos (t) + 1)

Solution steps

\int_{0}^{t} 2 sin (t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{t} sin (t) d t

Use the common integral:

\int sin (t) d t = - cos (t)

= 2 [- cos (t)]_{0}^{t}

Compute the boundaries:

- cos (t) + 1

= 2 (- cos (t) + 1)

\int_{0}^{\frac{π}{2}} 2 sin^{2} (x) d x

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\int_{1}^{6} 2 π y (6 - y) d y

\int_{1}^{\infty} \frac{1}{x + 4} d x

What is the integral from 0 to t of 2sin(t) ?
The integral from 0 to t of 2sin(t) is 2(-cos(t)+1)