What is the integral from 1 to pi of cos(3t)+8t ?

The integral from 1 to pi of cos(3t)+8t is -1/3 sin(3)+4(pi^2-1)

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integral from 1 to pi of cos(3t)+8t

\int_{1}^{π} cos (3 t) + 8 t d t

- \frac{1}{3} sin (3) + 4 (π^{2} - 1)

Solution steps

\int_{1}^{π} cos (3 t) + 8 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_{1}^{π} cos (3 t) d t + \int_{1}^{π} 8 t d t

\int_{1}^{π} cos (3 t) d t = - \frac{1}{3} sin (3)

\int_{1}^{π} 8 t d t = 4 (π^{2} - 1)

= - \frac{1}{3} sin (3) + 4 (π^{2} - 1)

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

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\int_{12}^{5} (12 - 6 x) d x

\int_{- 1}^{1} - u^{2} + 1 d u

What is the integral from 1 to pi of cos(3t)+8t ?
The integral from 1 to pi of cos(3t)+8t is -1/3 sin(3)+4(pi^2-1)