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

The integral from 0 to x^4 of cos^2(u) is 1/2 (x^4+1/2 sin(2x^4))

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integral from 0 to x^4 of cos^2(u)

\int_{0}^{x^{4}} cos^{2} (u) d u

\frac{1}{2} (x^{4} + \frac{1}{2} sin (2 x^{4}))

Solution steps

\int_{0}^{x^{4}} cos^{2} (u) d u

Rewrite using trig identities

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{x^{4}} 1 d u = x^{4}

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

= \frac{1}{2} (x^{4} + \frac{1}{2} sin (2 x^{4}))

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