What is the integral of (x^2)/(sqrt(196-x^2)) ?

The integral of (x^2)/(sqrt(196-x^2)) is 98(arcsin(1/14 x)-1/2 sin(2arcsin(1/14 x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(sqrt(196-x^2))

\int \frac{x ^{2}}{196 - x ^{2}} d x

98 (arcsin (\frac{1}{14} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{14} x))) + C

Solution steps

\int \frac{x ^{2}}{196 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 196 sin^{2} (u) d u

= \int 196 sin^{2} (u) d u

Take the constant out:

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

= 196 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 196 \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

= 196 \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

= 196 \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)

= 196 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{14} x)

= 196 \cdot \frac{1}{2} (arcsin (\frac{1}{14} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{14} x)))

Simplify

196 \cdot \frac{1}{2} (arcsin (\frac{1}{14} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{14} x))) : 98 (arcsin (\frac{1}{14} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{14} x)))

= 98 (arcsin (\frac{1}{14} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{14} x)))

Add a constant to the solution

= 98 (arcsin (\frac{1}{14} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{14} x))) + C

y^{^{''}} + 2 y^{^{'}} + 37 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

\frac{d}{d y} ((2 - y^{2} + x^{2}) e^{(1 - x^{2} - \frac{y ^{2}}{4})})

\frac{d}{d x} (36 e^{- 2 x})

f (x) = - 3 e^{4 x}

\int - 4 x^{- 5} d x

What is the integral of (x^2)/(sqrt(196-x^2)) ?
The integral of (x^2)/(sqrt(196-x^2)) is 98(arcsin(1/14 x)-1/2 sin(2arcsin(1/14 x)))+C