What is the integral from 0 to pi/2 of (x+xsin(x)) ?

The integral from 0 to pi/2 of (x+xsin(x)) is (pi^2)/8+1

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integral from 0 to pi/2 of (x+xsin(x))

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

\frac{π ^{2}}{8} + 1

Solution steps

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

\int_{0}^{\frac{π}{2}} x d x = \frac{π ^{2}}{8}

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

= \frac{π ^{2}}{8} + 1

\int_{\frac{π}{4}}^{\frac{π}{2}} cos^{2} (x) d x

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\int_{0}^{x} e^{- y} d y

What is the integral from 0 to pi/2 of (x+xsin(x)) ?
The integral from 0 to pi/2 of (x+xsin(x)) is (pi^2)/8+1