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

The integral from 0 to pi/2 of (1-sin(x)) is pi/2-1

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integral from 0 to pi/2 of (1-sin(x))

\int_{0}^{\frac{π}{2}} (1 - sin (x)) d x

\frac{π}{2} - 1

Decimal Notation

0.57079 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 1 - sin (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}^{\frac{π}{2}} 1 d x - \int_{0}^{\frac{π}{2}} sin (x) d x

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

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

= \frac{π}{2} - 1

\int_{- \infty}^{0} e^{- \frac{x}{2}} d x

\int_{0}^{2 π} cos (x^{2}) d x

\int_{0}^{4} 4 - x^{2} - (4 - 4 x) d x

\int_{1}^{4} - 2 x + 6 d x

\int_{- 1}^{2} \frac{1}{x ^{2} - 4 x + 13} d x

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