What is the integral of (u^2)/((9-u^2)^{3/2)} ?

The integral of (u^2)/((9-u^2)^{3/2)} is -arcsin(1/3 u)+u/((9-u^2)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (u^2)/((9-u^2)^{3/2)}

\int \frac{u ^{2}}{( 9 - u ^{2} ) ^{\frac{3}{2}}} d u

- arcsin (\frac{1}{3} u) + \frac{u}{( 9 - u ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{u ^{2}}{( 9 - u ^{2} ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int tan^{2} (v) d v

= \int tan^{2} (v) d v

Rewrite using trig identities

= \int - 1 + sec^{2} (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d v + \int sec^{2} (v) d v

\int 1 d v = v

\int sec^{2} (v) d v = tan (v)

= - v + tan (v)

Substitute back

v = arcsin (\frac{1}{3} u)

= - arcsin (\frac{1}{3} u) + tan (arcsin (\frac{1}{3} u))

Simplify

- arcsin (\frac{1}{3} u) + tan (arcsin (\frac{1}{3} u)) : - arcsin (\frac{1}{3} u) + \frac{u}{( 9 - u ^{2} ) ^{\frac{1}{2}}}

= - arcsin (\frac{1}{3} u) + \frac{u}{( 9 - u ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - arcsin (\frac{1}{3} u) + \frac{u}{( 9 - u ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{3 x - 1}{3 x} d x

\int 2 x cos (y) - y cos (x) d x

\int \frac{5 x}{2} d x

\int x sin (\frac{1}{11}) x d x

\int \frac{20 e ^{0.02 x}}{1 + e ^{0.02 x}} d x

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