What is the integral from 0 to 1 of arcsin(x)-sin(x) ?

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

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integral from 0 to 1 of arcsin(x)-sin(x)

\int_{0}^{1} arcsin (x) - sin (x) d x

\frac{π}{2} + cos (1) - 2

Solution steps

\int_{0}^{1} arcsin (x) - 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}^{1} arcsin (x) d x - \int_{0}^{1} sin (x) d x

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

\int_{0}^{1} sin (x) d x = - cos (1) + 1

= \frac{π}{2} - 1 - (- cos (1) + 1)

Simplify

= \frac{π}{2} + cos (1) - 2

\int (\frac{3 x ^{2} + 3 x + 1}{x}) d x

f (x) = \frac{x ^{2}}{x + 9}

\int \frac{1}{cos ^{2} ( 7 x )} d x

(6 x^{2} + y - 1) d x - (14 y - x) d y = 0, y (1) = 0

x^{2} \frac{d y}{d x} + 2 x y = 2 x

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