What is the integral of sqrt(x^3)sin^2(1+x^{5/2}) ?

The integral of sqrt(x^3)sin^2(1+x^{5/2}) is 1/10 (2+2x^{5/2}-sin(2(1+x^{5/2})))+C

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integral of sqrt(x^3)sin^2(1+x^{5/2})

\int x^{3} sin^{2} (1 + x^{\frac{5}{2}}) d x

\frac{1}{10} (2 + 2 x^{\frac{5}{2}} - sin (2 (1 + x^{\frac{5}{2}}))) + C

Solution steps

\int x^{3} sin^{2} (1 + x^{\frac{5}{2}}) d x

Apply u-substitution

= \int \frac{2 sin ^{2} ( u )}{5} d u

Take the constant out:

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

= \frac{2}{5} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= \frac{2}{5} \cdot \int \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{2}{5} \cdot \frac{1}{2} \cdot \int 1 - cos (2 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{2}{5} \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

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

= \frac{2}{5} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = 1 + x^{\frac{5}{2}}

= \frac{2}{5} \cdot \frac{1}{2} (1 + x^{\frac{5}{2}} - \frac{1}{2} sin (2 (1 + x^{\frac{5}{2}})))

Simplify

\frac{2}{5} \cdot \frac{1}{2} (1 + x^{\frac{5}{2}} - \frac{1}{2} sin (2 (1 + x^{\frac{5}{2}}))) : \frac{1}{10} (2 + 2 x^{\frac{5}{2}} - sin (2 (1 + x^{\frac{5}{2}})))

= \frac{1}{10} (2 + 2 x^{\frac{5}{2}} - sin (2 (1 + x^{\frac{5}{2}})))

Add a constant to the solution

= \frac{1}{10} (2 + 2 x^{\frac{5}{2}} - sin (2 (1 + x^{\frac{5}{2}}))) + C

What is the integral of sqrt(x^3)sin^2(1+x^{5/2}) ?
The integral of sqrt(x^3)sin^2(1+x^{5/2}) is 1/10 (2+2x^{5/2}-sin(2(1+x^{5/2})))+C