What is the integral from 0 to pi/3 of 49sin^2(x) ?

The integral from 0 to pi/3 of 49sin^2(x) is (49(4pi-3sqrt(3)))/(24)

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integral from 0 to pi/3 of 49sin^2(x)

\int_{0}^{\frac{π}{3}} 49 sin^{2} (x) d x

\frac{49 ( 4 π - 3 3 )}{24}

Decimal Notation

15.04752 \dots

Solution steps

\int_{0}^{\frac{π}{3}} 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}{3}} sin^{2} (x) d x

Rewrite using trig identities

= 49 \cdot \int_{0}^{\frac{π}{3}} \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}{3}} 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}{3}} 1 d x - \int_{0}^{\frac{π}{3}} cos (2 x) d x)

\int_{0}^{\frac{π}{3}} 1 d x = \frac{π}{3}

\int_{0}^{\frac{π}{3}} cos (2 x) d x = \frac{3}{4}

= 49 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

Simplify

49 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : \frac{49 ( 4 π - 3 3 )}{24}

= \frac{49 ( 4 π - 3 3 )}{24}

\int \frac{x ^{3} - 1}{( 2 x ^{4} - 8 x ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (2 cos (x) + cos^{2} (x))

\frac{\partial}{\partial x} (\frac{c ^{2} - x ^{2}}{2 x})

\int \frac{1}{( x ^{2} + 25 ) ^{\frac{5}{2}}} d x

f (t) = e^{t a n (π t)}

What is the integral from 0 to pi/3 of 49sin^2(x) ?
The integral from 0 to pi/3 of 49sin^2(x) is (49(4pi-3sqrt(3)))/(24)