What is the integral from 0 to 2pi of-4sin^2(t) ?

The integral from 0 to 2pi of-4sin^2(t) is -4pi

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integral from 0 to 2pi of-4sin^2(t)

\int_{0}^{2 π} - 4 sin^{2} (t) d t

- 4 π

Decimal Notation

- 12.56637 \dots

Solution steps

\int_{0}^{2 π} - 4 sin^{2} (t) d t

Take the constant out:

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

= - 4 \cdot \int_{0}^{2 π} sin^{2} (t) d t

Rewrite using trig identities

= - 4 \cdot \int_{0}^{2 π} \frac{1 - cos ( 2 t )}{2} d t

Take the constant out:

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

= - 4 \cdot \frac{1}{2} \cdot \int_{0}^{2 π} 1 - cos (2 t) d t

Apply the Sum Rule:

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

= - 4 \cdot \frac{1}{2} (\int_{0}^{2 π} 1 d t - \int_{0}^{2 π} cos (2 t) d t)

\int_{0}^{2 π} 1 d t = 2 π

\int_{0}^{2 π} cos (2 t) d t = 0

= - 4 \cdot \frac{1}{2} (2 π - 0)

Simplify

- 4 \cdot \frac{1}{2} (2 π - 0) : - 4 π

= - 4 π

\int_{0}^{3} x^{2} - 3 x + 2 d x

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

\int_{0}^{\frac{11}{12}} 20 x d x

\int_{2}^{\infty} \frac{1}{x ^{2} - x} d x

\int_{1}^{2} (6 x y^{2} - 4 \frac{x}{y}) d y

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