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

The integral of (sin^2(x))/(sqrt(x)) is 1/2 (2sqrt(x)-sqrt(pi)\C(sqrt(2)sqrt(2/pi)sqrt(x)))+C

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integral of (sin^2(x))/(sqrt(x))

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

\frac{1 - cos ( u )}{u} : \frac{1}{u} - \frac{cos ( u )}{u}

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

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

\int \frac{1}{u} d u = 2 u

\int \frac{cos ( u )}{u} d u = 2 π C (\frac{2}{π} u)

= \frac{1}{2} \cdot \frac{1}{2} (2 u - 2 π C (\frac{2}{π} u))

Substitute back

u = 2 x

= \frac{1}{2} \cdot \frac{1}{2} (2 2 x - 2 π C (\frac{2}{π} 2 x))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (2 2 x - 2 π C (\frac{2}{π} 2 x)) : \frac{1}{2} (2 x - π C (2 \frac{2}{π} x))

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

Add a constant to the solution

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

\int \frac{1}{2 x ^{3} + 2 x ^{2}} d x

\frac{d}{d x} (arcsin (3 x^{2}))

d er i v a t i v e - 2 y

t an g e n t f (x) = 5 e^{x} + x, (0, 5)

x \to b lim (\frac{x - b}{x - b})

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