What is the integral from 0 to pi of pi^{25}sin^2(x) ?

The integral from 0 to pi of pi^{25}sin^2(x) is (pi^{26})/2

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integral from 0 to pi of pi^{25}sin^2(x)

\int_{0}^{π} π^{25} sin^{2} (x) d x

\frac{π ^{26}}{2}

Decimal Notation

4215670845938.099

Solution steps

\int_{0}^{π} π^{25} sin^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π^{25} \cdot \int_{0}^{π} sin^{2} (x) d x

Rewrite using trig identities

= π^{25} \cdot \int_{0}^{π} \frac{1 - cos ( 2 x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π^{25} \frac{1}{2} \cdot \int_{0}^{π} 1 - cos (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

= π^{25} \frac{1}{2} (\int_{0}^{π} 1 d x - \int_{0}^{π} cos (2 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (2 x) d x = 0

= π^{25} \frac{1}{2} (π - 0)

Simplify

π^{25} \frac{1}{2} (π - 0) : \frac{π ^{26}}{2}

= \frac{π ^{26}}{2}

\int 1 d θ

\int_{0}^{9} 1 + x d x

n = 1 \sum \infty arctan (n^{2})

x \to 0 lim (\frac{( 2 ^{x} - 2 ^{- x} )}{x})

What is the integral from 0 to pi of pi^{25}sin^2(x) ?
The integral from 0 to pi of pi^{25}sin^2(x) is (pi^{26})/2