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

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

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integral of sin^2(1/4 x)

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

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

Solution steps

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

Rewrite using trig identities

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

\int 1 d x = x

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

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

Simplify

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

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

Add a constant to the solution

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

x \cdot \frac{d y}{d x} + y = - 6 x y^{2}, y (1) = - 7

\frac{d}{d x} (e^{(e^{x^{2}})})

\int \frac{e \frac{1}{t} + 2}{t ^{2}} d t

x \to 6 lim (\frac{x}{x ^{2} + 1})

x \to 1 - lim (\frac{1 + x}{1 - x})

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