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

The integral from 0 to pi of (cos(x)+4)^2 is 16pi+pi/2

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integral from 0 to pi of (cos(x)+4)^2

\int_{0}^{π} (cos (x) + 4)^{2} d x

16 π + \frac{π}{2}

Decimal Notation

51.83627 \dots

Solution steps

\int_{0}^{π} (cos (x) + 4)^{2} d x

Expand

(cos (x) + 4)^{2} : cos^{2} (x) + 8 cos (x) + 16

= \int_{0}^{π} cos^{2} (x) + 8 cos (x) + 16 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_{0}^{π} cos^{2} (x) d x + \int_{0}^{π} 8 cos (x) d x + \int_{0}^{π} 16 d x

\int_{0}^{π} cos^{2} (x) d x = \frac{π}{2}

\int_{0}^{π} 8 cos (x) d x = 0

\int_{0}^{π} 16 d x = 16 π

= \frac{π}{2} + 0 + 16 π

Simplify

= 16 π + \frac{π}{2}

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What is the integral from 0 to pi of (cos(x)+4)^2 ?
The integral from 0 to pi of (cos(x)+4)^2 is 16pi+pi/2