What is the integral of cos^4(1/2 x) ?

The integral of cos^4(1/2 x) is 2(cos^3(x/2)sin(x/2)+3/8 (x/2-1/4 sin(2x)))+C

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integral of cos^4(1/2 x)

\int cos^{4} (\frac{1}{2} x) d x

2 (cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} - \frac{1}{4} sin (2 x))) + C

Solution steps

\int cos^{4} (\frac{1}{2} x) d x

Simplify

\frac{1}{2} x : \frac{x}{2}

= \int cos^{4} (\frac{x}{2}) d x

Apply u-substitution

: \int cos^{4} (u) \cdot 2 d u

= \int cos^{4} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int cos^{4} (u) d u

Simplify

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

= 2 \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

= 2 (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))

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

Substitute back

u = \frac{x}{2}

= 2 (cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) - (- \frac{3}{8} (\frac{x}{2} - \frac{1}{4} sin (4 \cdot \frac{x}{2}))))

Simplify

2 (cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) - (- \frac{3}{8} (\frac{x}{2} - \frac{1}{4} sin (4 \cdot \frac{x}{2})))) : 2 (cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} - \frac{1}{4} sin (2 x)))

= 2 (cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} - \frac{1}{4} sin (2 x)))

Add a constant to the solution

= 2 (cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} - \frac{1}{4} sin (2 x))) + C

\int_{0}^{1} \frac{1}{2} x^{2} d x

x - 5 - 100

\frac{1}{2 x - 3}

\int θ 4 1 - θ^{2} d θ

\frac{1}{s ^{2} + 9}

What is the integral of cos^4(1/2 x) ?
The integral of cos^4(1/2 x) is 2(cos^3(x/2)sin(x/2)+3/8 (x/2-1/4 sin(2x)))+C