What is the integral from 0 to pi of (1+sin(x))^2 ?

The integral from 0 to pi of (1+sin(x))^2 is pi+4+pi/2

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integral from 0 to pi of (1+sin(x))^2

\int_{0}^{π} (1 + sin (x))^{2} d x

π + 4 + \frac{π}{2}

Decimal

8.71238 \dots

Solution steps

\int_{0}^{π} (1 + sin (x))^{2} d x

Expand

(1 + sin (x))^{2} : 1 + 2 sin (x) + sin^{2} (x)

= \int_{0}^{π} 1 + 2 sin (x) + sin^{2} (x) 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} 1 d x + \int_{0}^{π} 2 sin (x) d x + \int_{0}^{π} sin^{2} (x) d x

\int_{0}^{π} 1 d x = π

\int_{0}^{π} 2 sin (x) d x = 4

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

= π + 4 + \frac{π}{2}

What is the integral from 0 to pi of (1+sin(x))^2 ?
The integral from 0 to pi of (1+sin(x))^2 is pi+4+pi/2