What is the integral of \sqrt[5]{(2-aw^2)^3}w ?

The integral of \sqrt[5]{(2-aw^2)^3}w is -(5(2-aw^2)^{8/5})/(16a)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of \sqrt[5]{(2-aw^2)^3}w

- \frac{5 ( 2 - a w ^{2} ) ^{\frac{8}{5}}}{16 a} + C

Solution steps

Apply u-substitution

= \int - \frac{u ^{\frac{3}{5}}}{2 a} d u

Take the constant out:

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

= - \frac{1}{2 a} \cdot \int u^{\frac{3}{5}} d u

Apply the Power Rule

= - \frac{1}{2 a} \cdot \frac{5}{8} u^{\frac{8}{5}}

Substitute back

u = 2 - a w^{2}

= - \frac{1}{2 a} \cdot \frac{5}{8} (2 - a w^{2})^{\frac{8}{5}}

Simplify

- \frac{1}{2 a} \cdot \frac{5}{8} (2 - a w^{2})^{\frac{8}{5}} : - \frac{5 ( 2 - a w ^{2} ) ^{\frac{8}{5}}}{16 a}

= - \frac{5 ( 2 - a w ^{2} ) ^{\frac{8}{5}}}{16 a}

Add a constant to the solution

= - \frac{5 ( 2 - a w ^{2} ) ^{\frac{8}{5}}}{16 a} + C

What is the integral of \sqrt[5]{(2-aw^2)^3}w ?
The integral of \sqrt[5]{(2-aw^2)^3}w is -(5(2-aw^2)^{8/5})/(16a)+C