What is the integral from 0 to pi/4 of 49sin^2(x) ?

The integral from 0 to pi/4 of 49sin^2(x) is (49(pi-2))/8

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integral from 0 to pi/4 of 49sin^2(x)

\int_{0}^{\frac{π}{4}} 49 sin^{2} (x) d x

\frac{49 ( π - 2 )}{8}

Decimal Notation

6.99225 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 49 sin^{2} (x) d x

Take the constant out:

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

= 49 \cdot \int_{0}^{\frac{π}{4}} sin^{2} (x) d x

Rewrite using trig identities

= 49 \cdot \int_{0}^{\frac{π}{4}} \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

= 49 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 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

= 49 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d x - \int_{0}^{\frac{π}{4}} cos (2 x) d x)

\int_{0}^{\frac{π}{4}} 1 d x = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 x) d x = \frac{1}{2}

= 49 \cdot \frac{1}{2} (\frac{π}{4} - \frac{1}{2})

Simplify

49 \cdot \frac{1}{2} (\frac{π}{4} - \frac{1}{2}) : \frac{49 ( π - 2 )}{8}

= \frac{49 ( π - 2 )}{8}

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What is the integral from 0 to pi/4 of 49sin^2(x) ?
The integral from 0 to pi/4 of 49sin^2(x) is (49(pi-2))/8