What is the integral from-2pi to 2pi of (sin^2(x)+1) ?

The integral from-2pi to 2pi of (sin^2(x)+1) is 6pi

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integral from-2pi to 2pi of (sin^2(x)+1)

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

6 π

Decimal Notation

18.84955 \dots

Solution steps

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

Apply the Sum Rule:

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

= \int_{- 2 π}^{2 π} sin^{2} (x) d x + \int_{- 2 π}^{2 π} 1 d x

\int_{- 2 π}^{2 π} sin^{2} (x) d x = 2 π

\int_{- 2 π}^{2 π} 1 d x = 4 π

= 2 π + 4 π

Simplify

= 6 π

\int lo g_{e} (5 x) d x

f (x) = - 2 sin (x)

300 - 3 p

\frac{\partial}{\partial x} (arcsin (y))

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What is the integral from-2pi to 2pi of (sin^2(x)+1) ?
The integral from-2pi to 2pi of (sin^2(x)+1) is 6pi