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

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

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integral from 0 to x of sin(2t)

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

sin^{2} (x)

Solution steps

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

Apply u-substitution

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

= \int_{0}^{2 x} 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 x} sin (u) d u

Use the common integral:

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

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

Compute the boundaries:

2 sin^{2} (x)

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

Simplify

= sin^{2} (x)

\int_{- 1}^{4} \frac{3}{2 x ^{2}} d x

\int_{0}^{2} \frac{5 x ^{2} + 1}{x ^{3} + 1} d x

\int_{0}^{4} sec (t) tan (t) d t

\int_{0}^{2} - x d x

\int_{3}^{- 3} (6 r + 5)^{2} d r

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