What is the integral of-2x^3sqrt(9-x^2) ?

The integral of-2x^3sqrt(9-x^2) is 2(-1/5 (9-x^2)^{5/2}+3(9-x^2)^{3/2})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-2x^3sqrt(9-x^2)

\int - 2 x^{3} 9 - x^{2} d x

2 (- \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} + 3 (9 - x^{2})^{\frac{3}{2}}) + C

Solution steps

\int - 2 x^{3} 9 - 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^{3} 9 - x^{2} d x

Apply u-substitution

: \int - (- u^{2} + 9) u^{2} d u

= - 2 \cdot \int - (- u^{2} + 9) 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} + 9) u^{2} d u)

Expand

(- u^{2} + 9) u^{2} : - u^{4} + 9 u^{2}

= - 2 (- \int - u^{4} + 9 u^{2} 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

= - 2 (- (- \int u^{4} d u + \int 9 u^{2} d u))

\int u^{4} d u = \frac{u ^{5}}{5}

\int 9 u^{2} d u = 3 u^{3}

= - 2 (- (- \frac{u ^{5}}{5} + 3 u^{3}))

Substitute back

u = 9 - x^{2}

= - 2 (- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3}))

Simplify

- 2 (- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3})) : 2 (- \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} + 3 (9 - x^{2})^{\frac{3}{2}})

= 2 (- \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} + 3 (9 - x^{2})^{\frac{3}{2}})

Add a constant to the solution

= 2 (- \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} + 3 (9 - x^{2})^{\frac{3}{2}}) + C

\int x d ln (x + 1 + x^{2}) d x

\int (5 x^{4} + 6 x + 3) d x

\int - \frac{cos ( x )}{sin ^{4} ( x )} d x

\int x (x^{2} - 2)^{15} d x

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

What is the integral of-2x^3sqrt(9-x^2) ?
The integral of-2x^3sqrt(9-x^2) is 2(-1/5 (9-x^2)^{5/2}+3(9-x^2)^{3/2})+C