What is the integral from-7 to 5 of sin(2x)cos(2x) ?

The integral from-7 to 5 of sin(2x)cos(2x) is 1/8 (-cos(20)+cos(28))

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integral from-7 to 5 of sin(2x)cos(2x)

\int_{- 7}^{5} sin (2 x) cos (2 x) d x

\frac{1}{8} (- cos (20) + cos (28))

Decimal

- 0.17133 \dots

Solution steps

\int_{- 7}^{5} sin (2 x) cos (2 x) d x

Rewrite using trig identities

= \int_{- 7}^{5} \frac{1}{2} sin (4 x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 7}^{5} sin (4 x) d x

Apply u-substitution

: \int_{- 28}^{20} sin (u) \frac{1}{4} d u

= \frac{1}{2} \cdot \int_{- 28}^{20} sin (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int_{- 28}^{20} sin (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{4} [- cos (u)]_{- 28}^{20}

Simplify

\frac{1}{2} \cdot \frac{1}{4} [- cos (u)]_{- 28}^{20} : \frac{1}{8} [- cos (u)]_{- 28}^{20}

= \frac{1}{8} [- cos (u)]_{- 28}^{20}

Compute the boundaries:

- cos (20) + cos (28)

= \frac{1}{8} (- cos (20) + cos (28))

\int_{0}^{310} x^{2} d x

\int_{- 1}^{1} (- 30 x^{2} + 4 x y) d y

\int_{0}^{5} t^{2} + 2 t - 15 d t

\int_{0}^{2} \frac{1}{( 2 - x ) ^{\frac{5}{2}}} d x

\int_{2}^{4} (2 x) d x

What is the integral from-7 to 5 of sin(2x)cos(2x) ?
The integral from-7 to 5 of sin(2x)cos(2x) is 1/8 (-cos(20)+cos(28))