What is the integral from 0 to 4 of 1/2 (4cos(5θ))^2 ?

The integral from 0 to 4 of 1/2 (4cos(5θ))^2 is 1/2 (4/5 sin(40)+32)

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integral from 0 to 4 of 1/2 (4cos(5θ))^2

\int_{0}^{4} \frac{1}{2} (4 cos (5 θ))^{2} d θ

\frac{1}{2} (\frac{4}{5} sin (40) + 32)

Decimal

16.29804 \dots

Solution steps

\int_{0}^{4} \frac{1}{2} (4 cos (5 θ))^{2} d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{4} (4 cos (5 θ))^{2} d θ

Apply Integration By Parts

= \frac{1}{2} [16 θ cos^{2} (5 θ) - \int - 160 θ cos (5 θ) sin (5 θ) d θ]_{0}^{4}

\int - 160 θ cos (5 θ) sin (5 θ) d θ = - 160 (- \frac{1}{10} θ cos^{2} (5 θ) + \frac{1}{20} (θ + \frac{1}{10} sin (10 θ)))

= \frac{1}{2} [16 θ cos^{2} (5 θ) - (- 160 (- \frac{1}{10} θ cos^{2} (5 θ) + \frac{1}{20} (θ + \frac{1}{10} sin (10 θ))))]_{0}^{4}

Simplify

\frac{1}{2} [16 θ cos^{2} (5 θ) - (- 160 (- \frac{1}{10} θ cos^{2} (5 θ) + \frac{1}{20} (θ + \frac{1}{10} sin (10 θ))))]_{0}^{4} : \frac{1}{2} [8 θ + \frac{4}{5} sin (10 θ)]_{0}^{4}

= \frac{1}{2} [8 θ + \frac{4}{5} sin (10 θ)]_{0}^{4}

Compute the boundaries:

\frac{4}{5} sin (40) + 32

= \frac{1}{2} (\frac{4}{5} sin (40) + 32)

What is the integral from 0 to 4 of 1/2 (4cos(5θ))^2 ?
The integral from 0 to 4 of 1/2 (4cos(5θ))^2 is 1/2 (4/5 sin(40)+32)