What is the integral from 0 to 2pi of-3sin^2(t) ?

The integral from 0 to 2pi of-3sin^2(t) is -3pi

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integral from 0 to 2pi of-3sin^2(t)

\int_{0}^{2 π} - 3 sin^{2} (t) d t

- 3 π

Decimal

- 9.42477 \dots

Solution steps

\int_{0}^{2 π} - 3 sin^{2} (t) d t

Take the constant out:

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

= - 3 \cdot \int_{0}^{2 π} sin^{2} (t) d t

Rewrite using trig identities

= - 3 \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

= - 3 \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

= - 3 \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

= - 3 \cdot \frac{1}{2} (2 π - 0)

Simplify

- 3 \cdot \frac{1}{2} (2 π - 0) : - 3 π

= - 3 π

\int_{x}^{\infty} e^{- x y} d y

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

\int_{0}^{1} e^{8 x - 7} d x

\int_{0}^{2} \frac{3}{4} x (2 x - x^{2}) d x

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

What is the integral from 0 to 2pi of-3sin^2(t) ?
The integral from 0 to 2pi of-3sin^2(t) is -3pi