What is the integral from 0 to 2pi of-16sin^2(θ) ?

The integral from 0 to 2pi of-16sin^2(θ) is -16pi

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integral from 0 to 2pi of-16sin^2(θ)

\int_{0}^{2 π} - 16 sin^{2} (θ) d θ

- 16 π

Decimal

- 50.26548 \dots

Solution steps

\int_{0}^{2 π} - 16 sin^{2} (θ) d θ

Take the constant out:

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

= - 16 \cdot \int_{0}^{2 π} sin^{2} (θ) d θ

Rewrite using trig identities

= - 16 \cdot \int_{0}^{2 π} \frac{1 - cos ( 2 θ )}{2} d θ

Take the constant out:

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

= - 16 \cdot \frac{1}{2} \cdot \int_{0}^{2 π} 1 - cos (2 θ) d θ

Apply the Sum Rule:

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

= - 16 \cdot \frac{1}{2} (\int_{0}^{2 π} 1 d θ - \int_{0}^{2 π} cos (2 θ) d θ)

\int_{0}^{2 π} 1 d θ = 2 π

\int_{0}^{2 π} cos (2 θ) d θ = 0

= - 16 \cdot \frac{1}{2} (2 π - 0)

Simplify

- 16 \cdot \frac{1}{2} (2 π - 0) : - 16 π

= - 16 π

\int_{0}^{1} 2 y e^{- y^{2}} d y

\int_{0}^{2} x + y d y

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

\int_{0}^{3} \frac{1}{4} e^{- \frac{x}{4}} d x

\int_{0}^{+} 1

What is the integral from 0 to 2pi of-16sin^2(θ) ?
The integral from 0 to 2pi of-16sin^2(θ) is -16pi