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

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_{1}^{2} \frac{e ^{\frac{1}{x}}}{x} d x

\int_{2}^{4} e^{x^{2}} d x

\int_{0}^{1} \frac{1}{( x + 1 ) ^{5}} d x

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

\int_{0}^{4} x^{2} - 9 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