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

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

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integral of sin^2(x/4)

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

\frac{1}{2} x - sin (\frac{x}{2}) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (\frac{x}{2})) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 - cos (\frac{x}{2}) d x

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

\int 1 d x = x

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

= \frac{1}{2} (x - 2 sin (\frac{x}{2}))

Simplify

\frac{1}{2} (x - 2 sin (\frac{x}{2})) : \frac{1}{2} x - sin (\frac{x}{2})

= \frac{1}{2} x - sin (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} x - sin (\frac{x}{2}) + C

y^{^{'}} + x y = x y^{- 1}

\frac{d}{d x} (x ey (x)^{2} + y ln (x))

\int \frac{1}{x ^{4} 1 + x ^{2}} d x

ln (1 - 3 y)

\int \frac{( x ^{2} )}{x ^{2} + 9} d x

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