What is the integral of cos^4(x/4) ?

The integral of cos^4(x/4) is 4(cos^3(x/4)sin(x/4)+3/32 (x-sin(x)))+C

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integral of cos^4(x/4)

\int cos^{4} (\frac{x}{4}) d x

4 (cos^{3} (\frac{x}{4}) sin (\frac{x}{4}) + \frac{3}{32} (x - sin (x))) + C

Solution steps

\int cos^{4} (\frac{x}{4}) d x

Apply u-substitution

: \int cos^{4} (u) \cdot 4 d u

= \int cos^{4} (u) \cdot 4 d u

Take the constant out:

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

= 4 \cdot \int cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= 4 \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

: cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u

= 4 (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= 4 (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = \frac{x}{4}

= 4 (cos^{3} (\frac{x}{4}) sin (\frac{x}{4}) - (- \frac{3}{8} (\frac{x}{4} - \frac{1}{4} sin (4 \cdot \frac{x}{4}))))

Simplify

4 (cos^{3} (\frac{x}{4}) sin (\frac{x}{4}) - (- \frac{3}{8} (\frac{x}{4} - \frac{1}{4} sin (4 \cdot \frac{x}{4})))) : 4 (cos^{3} (\frac{x}{4}) sin (\frac{x}{4}) + \frac{3}{32} (x - sin (x)))

= 4 (cos^{3} (\frac{x}{4}) sin (\frac{x}{4}) + \frac{3}{32} (x - sin (x)))

Add a constant to the solution

= 4 (cos^{3} (\frac{x}{4}) sin (\frac{x}{4}) + \frac{3}{32} (x - sin (x))) + C

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What is the integral of cos^4(x/4) ?
The integral of cos^4(x/4) is 4(cos^3(x/4)sin(x/4)+3/32 (x-sin(x)))+C