What is the integral of 2xsqrt(1-x^2) ?

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

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integral of 2xsqrt(1-x^2)

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

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

Solution steps

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

Take the constant out:

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

= 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

\int \frac{1}{- x ^{2} - 8 x - 7} d x

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

\frac{\partial}{\partial x} (y^{1 - x} ln^{2} (z))

\frac{d}{d x} (c \cdot ln (x))

x \to 0 + lim (0 \cdot sin (\frac{1}{x}))

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