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

The integral from 0 to pi/2 of y+y^2cos(x) is (piy)/2+y^2

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

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

\frac{π y}{2} + y^{2}

Solution steps

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

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

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

= \frac{π y}{2} + y^{2}

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

\int_{0}^{1} \frac{1}{2 + cos ( π x )} d x

\int_{- π}^{0} (t + \frac{π}{2}) cos (n t) d t

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

\int_{- 4}^{- 1} (x^{3} + 2 x^{2} + 1) d x

What is the integral from 0 to pi/2 of y+y^2cos(x) ?
The integral from 0 to pi/2 of y+y^2cos(x) is (piy)/2+y^2