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

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

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integral from 0 to pi/2 of x-cos(y)

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

- 1 + \frac{π x}{2}

Solution steps

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

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

\int_{0}^{\frac{π}{2}} cos (y) d y = 1

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

= - 1 + \frac{π x}{2}

\int_{0}^{6} 3 + 2 cos (\frac{π x}{6}) d x

\int_{0}^{2} e^{x^{4}} x^{3} d x

\int_{- 2}^{0} 2 + \frac{x ^{3}}{4} d x

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

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