What is the integral of (4-x^2)^{-3/2} ?

The integral of (4-x^2)^{-3/2} is x/(4(4-x^2)^{1/2)}+C

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integral of (4-x^2)^{-3/2}

\int (4 - x^{2})^{- \frac{3}{2}} d x

\frac{x}{4 ( 4 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int (4 - x^{2})^{- \frac{3}{2}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{4 cos ^{2} ( u )} d u

= \int \frac{1}{4 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}{4} \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}{4} tan (u)

Substitute back

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

= \frac{1}{4} tan (arcsin (\frac{1}{2} x))

Simplify

\frac{1}{4} tan (arcsin (\frac{1}{2} x)) : \frac{x}{4 ( 4 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{4 ( 4 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{4 ( 4 - x ^{2} ) ^{\frac{1}{2}}} + C

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What is the integral of (4-x^2)^{-3/2} ?
The integral of (4-x^2)^{-3/2} is x/(4(4-x^2)^{1/2)}+C