What is the integral from 0 to pi/2 of 1-4sin^2(x) ?

The integral from 0 to pi/2 of 1-4sin^2(x) is pi/2-pi

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integral from 0 to pi/2 of 1-4sin^2(x)

\int_{0}^{\frac{π}{2}} 1 - 4 sin^{2} (x) d x

\frac{π}{2} - π

Decimal Notation

- 1.57079 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 1 - 4 sin^{2} (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}} 1 d x - \int_{0}^{\frac{π}{2}} 4 sin^{2} (x) d x

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

\int_{0}^{\frac{π}{2}} 4 sin^{2} (x) d x = π

= \frac{π}{2} - π

2 x^{- \frac{3}{4}}

\frac{d y}{d x} = (1 + e^{x^{2}}) (5 + x)

\frac{4}{cos ( x )} + \frac{1}{tan ( x )}

\frac{d}{d x} (x e^{(- x^{2})})

\frac{( 4 x ^{2} + 8 x + 2 )}{x}

What is the integral from 0 to pi/2 of 1-4sin^2(x) ?
The integral from 0 to pi/2 of 1-4sin^2(x) is pi/2-pi