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

The integral of sqrt(x)*sin^2(x^{3/2}-1) is 1/6 (2x^{3/2}-2-sin(2(x^{3/2}-1)))+C

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integral of sqrt(x)*sin^2(x^{3/2}-1)

\int x \cdot sin^{2} (x^{\frac{3}{2}} - 1) d x

\frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1))) + C

Solution steps

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

Apply u-substitution

= \int \frac{2 sin ^{2} ( u )}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= \frac{2}{3} \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}{3} \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}{3} \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}{3} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = x^{\frac{3}{2}} - 1

= \frac{2}{3} \cdot \frac{1}{2} (x^{\frac{3}{2}} - 1 - \frac{1}{2} sin (2 (x^{\frac{3}{2}} - 1)))

Simplify

\frac{2}{3} \cdot \frac{1}{2} (x^{\frac{3}{2}} - 1 - \frac{1}{2} sin (2 (x^{\frac{3}{2}} - 1))) : \frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1)))

= \frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1)))

Add a constant to the solution

= \frac{1}{6} (2 x^{\frac{3}{2}} - 2 - sin (2 (x^{\frac{3}{2}} - 1))) + C

What is the integral of sqrt(x)*sin^2(x^{3/2}-1) ?
The integral of sqrt(x)*sin^2(x^{3/2}-1) is 1/6 (2x^{3/2}-2-sin(2(x^{3/2}-1)))+C