What is the integral from 0 to (3pi)/2 of cos(2x) ?

The integral from 0 to (3pi)/2 of cos(2x) is 0

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integral from 0 to (3pi)/2 of cos(2x)

\int_{0}^{\frac{3 π}{2}} cos (2 x) d x

0

Solution steps

\int_{0}^{\frac{3 π}{2}} cos (2 x) d x

Apply u-substitution

: \int_{0}^{3 π} cos (u) \frac{1}{2} d u

= \int_{0}^{3 π} cos (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}^{3 π} cos (u) d u

Use the common integral:

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

= \frac{1}{2} [sin (u)]_{0}^{3 π}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

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\int_{- 1}^{2} y^{2} - y - 2 d y

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

What is the integral from 0 to (3pi)/2 of cos(2x) ?
The integral from 0 to (3pi)/2 of cos(2x) is 0