What is the integral of cos^6(x/2) ?

The integral of cos^6(x/2) is 2(1/6 cos^5(x/2)sin(x/2)+5/96 (4cos^3(x/2)sin(x/2)+3(x+sin(x))))+C

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integral of cos^6(x/2)

\int cos^{6} (\frac{x}{2}) d x

2 (\frac{1}{6} cos^{5} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{5}{96} (4 cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + 3 (x + sin (x)))) + C

Solution steps

\int cos^{6} (\frac{x}{2}) d x

Apply u-substitution

: \int cos^{6} (u) \cdot 2 d u

= \int cos^{6} (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^{6} (u) d u

Apply Integral Reduction

= 2 (\frac{sin ( u ) cos ^{5} ( u )}{6} + \frac{5}{6} \cdot \int cos^{4} (u) d u)

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

= 2 (\frac{sin ( u ) cos ^{5} ( u )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (u) sin (u) + \frac{3}{8} (u + \frac{1}{2} sin (2 u))))

Substitute back

u = \frac{x}{2}

= 2 (\frac{sin ( \frac{x}{2} ) cos ^{5} ( \frac{x}{2} )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} + \frac{1}{2} sin (2 \cdot \frac{x}{2}))))

Simplify

2 (\frac{sin ( \frac{x}{2} ) cos ^{5} ( \frac{x}{2} )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} + \frac{1}{2} sin (2 \cdot \frac{x}{2})))) : 2 (\frac{1}{6} cos^{5} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{5}{96} (4 cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + 3 (x + sin (x))))

= 2 (\frac{1}{6} cos^{5} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{5}{96} (4 cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + 3 (x + sin (x))))

Add a constant to the solution

= 2 (\frac{1}{6} cos^{5} (\frac{x}{2}) sin (\frac{x}{2}) + \frac{5}{96} (4 cos^{3} (\frac{x}{2}) sin (\frac{x}{2}) + 3 (x + sin (x)))) + C

\int 9 e^{- t} d t

x \to \infty lim (- 1 \cdot x)

f (x) = sec^{2} (x) - tan^{2} (x)

\int e^{2 x} \cdot x d x

What is the integral of cos^6(x/2) ?
The integral of cos^6(x/2) is 2(1/6 cos^5(x/2)sin(x/2)+5/96 (4cos^3(x/2)sin(x/2)+3(x+sin(x))))+C