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}^{2 π} e^{s i n (x)} d x

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

\int_{\frac{π}{4}}^{\frac{π}{2}} 9 cot^{3} (x) d x

\int_{0}^{1} (e^{x} + 1) (e^{x} - 1) d x

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