What is the integral from 0 to pi/4 of 16cos^2(x) ?

The integral from 0 to pi/4 of 16cos^2(x) is 8(pi/4+1/2)

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integral from 0 to pi/4 of 16cos^2(x)

\int_{0}^{\frac{π}{4}} 16 cos^{2} (x) d x

8 (\frac{π}{4} + \frac{1}{2})

Decimal

10.28318 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 16 cos^{2} (x) d x

Take the constant out:

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

= 16 \cdot \int_{0}^{\frac{π}{4}} cos^{2} (x) d x

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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}

= 16 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

16 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : 8 (\frac{π}{4} + \frac{1}{2})

= 8 (\frac{π}{4} + \frac{1}{2})

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What is the integral from 0 to pi/4 of 16cos^2(x) ?
The integral from 0 to pi/4 of 16cos^2(x) is 8(pi/4+1/2)