What is the integral from 0 to 2pi of-36sin^2(t) ?

The integral from 0 to 2pi of-36sin^2(t) is -36pi

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integral from 0 to 2pi of-36sin^2(t)

\int_{0}^{2 π} - 36 sin^{2} (t) d t

- 36 π

Decimal

- 113.09733 \dots

Solution steps

\int_{0}^{2 π} - 36 sin^{2} (t) d t

Take the constant out:

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

= - 36 \cdot \int_{0}^{2 π} sin^{2} (t) d t

Rewrite using trig identities

= - 36 \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

= - 36 \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

= - 36 \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

= - 36 \cdot \frac{1}{2} (2 π - 0)

Simplify

- 36 \cdot \frac{1}{2} (2 π - 0) : - 36 π

= - 36 π

\int_{0}^{2} 36 t^{2} + 9 t^{4} d t

\int_{0}^{\infty} 1 - e^{- 2 x} d x

\int_{- x}^{x} sin (x) d x

\int_{- \infty}^{0} \frac{2}{x ^{2} + 4} d x

\int_{- 1}^{2} (x^{2} + 1) (x - 3) d x

What is the integral from 0 to 2pi of-36sin^2(t) ?
The integral from 0 to 2pi of-36sin^2(t) is -36pi