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

The integral from 0 to 1 of-2sin^2(2-2t) is -1+1/4 sin(4)

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integral from 0 to 1 of-2sin^2(2-2t)

\int_{0}^{1} - 2 sin^{2} (2 - 2 t) d t

- 1 + \frac{1}{4} sin (4)

Solution steps

\int_{0}^{1} - 2 sin^{2} (2 - 2 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}^{1} sin^{2} (2 - 2 t) d t

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Rewrite using trig identities

= - 2 (- (- \frac{1}{2} \cdot \int_{0}^{2} \frac{1 - cos ( 2 u )}{2} d u))

Take the constant out:

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

= - 2 (- (- \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{2} 1 - cos (2 u) d u))

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 2 (- (- \frac{1}{2} \cdot \frac{1}{2} (\int_{0}^{2} 1 d u - \int_{0}^{2} cos (2 u) d u)))

\int_{0}^{2} 1 d u = 2

\int_{0}^{2} cos (2 u) d u = \frac{1}{2} sin (4)

= - 2 (- (- \frac{1}{2} \cdot \frac{1}{2} (2 - \frac{1}{2} sin (4))))

Simplify

- 2 (- (- \frac{1}{2} \cdot \frac{1}{2} (2 - \frac{1}{2} sin (4)))) : - 1 + \frac{1}{4} sin (4)

= - 1 + \frac{1}{4} sin (4)

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\int_{0}^{5} (75 - 3 x^{2}) d x

What is the integral from 0 to 1 of-2sin^2(2-2t) ?
The integral from 0 to 1 of-2sin^2(2-2t) is -1+1/4 sin(4)