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

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

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

\int_{0}^{2 π} - 4 cos^{2} (t) d t

- 4 π

Decimal

- 12.56637 \dots

Solution steps

\int_{0}^{2 π} - 4 cos^{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} cos^{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_{3}^{5} \frac{x}{x ^{2} + 4 x + 13} d x

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

\int_{2}^{4} x^{2} + 2 x d x

\int_{- 3}^{1} x^{2} d x

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

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