What is the integral of sin(2arcsin(x)) ?

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

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integral of sin(2arcsin(x))

\int sin (2 arcsin (x)) d x

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

Solution steps

\int sin (2 arcsin (x)) d x

Rewrite using trig identities

= \int 2 cos (arcsin (x)) sin (arcsin (x)) d x

Take the constant out:

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

= 2 \cdot \int cos (arcsin (x)) sin (arcsin (x)) d x

Simplify

cos (arcsin (x)) sin (arcsin (x)) : x 1 - x^{2}

= 2 \cdot \int x 1 - x^{2} d x

Apply u-substitution

: \int - u^{2} d u

= 2 \cdot \int - u^{2} d u

Take the constant out:

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

= 2 (- \int u^{2} d u)

Apply the Power Rule

: \frac{u ^{3}}{3}

= 2 (- \frac{u ^{3}}{3})

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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