What is the integral from 0 to 2pi of-6sin^2(t) ?

The integral from 0 to 2pi of-6sin^2(t) is -6pi

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integral from 0 to 2pi of-6sin^2(t)

\int_{0}^{2 π} - 6 sin^{2} (t) d t

- 6 π

Decimal

- 18.84955 \dots

Solution steps

\int_{0}^{2 π} - 6 sin^{2} (t) d t

Take the constant out:

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

= - 6 \cdot \int_{0}^{2 π} sin^{2} (t) d t

Rewrite using trig identities

= - 6 \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

= - 6 \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

= - 6 \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

= - 6 \cdot \frac{1}{2} (2 π - 0)

Simplify

- 6 \cdot \frac{1}{2} (2 π - 0) : - 6 π

= - 6 π

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

\int_{0}^{1} 2 + 5 x d x

\int_{- \frac{π}{4}}^{0} sec (x) tan (x) d x

\int_{1}^{5} 7 - 2 x d x

\int_{1}^{4} (1 + \frac{1}{x} + \frac{1}{x ^{2}}) d x

What is the integral from 0 to 2pi of-6sin^2(t) ?
The integral from 0 to 2pi of-6sin^2(t) is -6pi