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

The integral from 0 to t of sin(2t) is sin^2(t)

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integral from 0 to t of sin(2t)

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

sin^{2} (t)

Solution steps

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

Apply u-substitution

: \int_{0}^{2 t} sin (u) \frac{1}{2} d u

= \int_{0}^{2 t} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 t} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{2 t}

Compute the boundaries:

2 sin^{2} (t)

= \frac{1}{2} \cdot 2 sin^{2} (t)

Simplify

= sin^{2} (t)

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\int_{0}^{1} 2^{4 x} d x

\int_{- 2}^{2} 5 d x

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