What is the integral from 0 to 2pi of-3sin^2(t)+9t ?

The integral from 0 to 2pi of-3sin^2(t)+9t is -3pi+18pi^2

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integral from 0 to 2pi of-3sin^2(t)+9t

\int_{0}^{2 π} - 3 sin^{2} (t) + 9 t d t

- 3 π + 18 π^{2}

Decimal

168.22810 \dots

Solution steps

\int_{0}^{2 π} - 3 sin^{2} (t) + 9 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

= - \int_{0}^{2 π} 3 sin^{2} (t) d t + \int_{0}^{2 π} 9 t d t

\int_{0}^{2 π} 3 sin^{2} (t) d t = 3 π

\int_{0}^{2 π} 9 t d t = 18 π^{2}

= - 3 π + 18 π^{2}

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What is the integral from 0 to 2pi of-3sin^2(t)+9t ?
The integral from 0 to 2pi of-3sin^2(t)+9t is -3pi+18pi^2