What is the integral of (w^2)/(sqrt(16-w^2)) ?

The integral of (w^2)/(sqrt(16-w^2)) is 8(arcsin(1/4 w)-1/2 sin(2arcsin(1/4 w)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (w^2)/(sqrt(16-w^2))

\int \frac{w ^{2}}{16 - w ^{2}} d w

8 (arcsin (\frac{1}{4} w) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} w))) + C

Solution steps

\int \frac{w ^{2}}{16 - w ^{2}} d w

Apply Trigonometric Substitution

= \int 16 sin^{2} (u) d u

Take the constant out:

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

= 16 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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)

= 16 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{4} w)

= 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} w) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} w)))

Simplify

16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} w) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} w))) : 8 (arcsin (\frac{1}{4} w) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} w)))

= 8 (arcsin (\frac{1}{4} w) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} w)))

Add a constant to the solution

= 8 (arcsin (\frac{1}{4} w) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} w))) + C

What is the integral of (w^2)/(sqrt(16-w^2)) ?
The integral of (w^2)/(sqrt(16-w^2)) is 8(arcsin(1/4 w)-1/2 sin(2arcsin(1/4 w)))+C