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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec^{2} (u) d u

= \int sec^{2} (u) d u

Use the common integral:

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

= tan (u)

Substitute back

u = arcsin (x)

= tan (arcsin (x))

Use the following identity:

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

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

Add a constant to the solution

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

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\frac{\partial}{\partial x} (sin (π (3 x - 2 y)))

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

\int 6 cos (x) d x

\int_{0}^{1} \frac{5}{x ^{2} + x - 6} d x

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