What is the integral of 1/((2-x^2)^{3/2)} ?

The integral of 1/((2-x^2)^{3/2)} is x/(2(2-x^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 1/((2-x^2)^{3/2)}

\int \frac{1}{( 2 - x ^{2} ) ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{1}{( 2 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{2 cos ^{2} ( u )} d u

= \int \frac{1}{2 cos ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{2} tan (u)

Substitute back

u = arcsin (\frac{1}{2 ^{\frac{1}{2}}} x)

= \frac{1}{2} tan (arcsin (\frac{1}{2 ^{\frac{1}{2}}} x))

Simplify

\frac{1}{2} tan (arcsin (\frac{1}{2 ^{\frac{1}{2}}} x)) : \frac{x}{2 ( 2 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{2 ( 2 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

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

\int - 2 x^{4} + 3 x^{2} - 1 d x

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

\int \frac{sin ( t )}{1 + sin ( t )} d t

\int cos (3 x + π) d x

\int sin^{7} (\frac{x}{3}) cos (\frac{x}{3}) d x

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