What is the integral of 1/((225-x^2)^{3/2)} ?

The integral of 1/((225-x^2)^{3/2)} is x/(225(225-x^2)^{1/2)}+C

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integral of 1/((225-x^2)^{3/2)}

\int \frac{1}{( 225 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{225 ( 225 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 225 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{225 cos ^{2} ( u )} d u

= \int \frac{1}{225 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}{225} \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}{225} tan (u)

Substitute back

u = arcsin (\frac{1}{15} x)

= \frac{1}{225} tan (arcsin (\frac{1}{15} x))

Simplify

\frac{1}{225} tan (arcsin (\frac{1}{15} x)) : \frac{x}{225 ( 225 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{225 ( 225 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{225 ( 225 - x ^{2} ) ^{\frac{1}{2}}} + C

(y^{2} sin (x)) d x + (\frac{1}{x} - \frac{y}{x}) d y = 0

x \to 4 lim (\frac{8 - 2 x}{x - 4})

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

y = \frac{d y}{d x}

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

What is the integral of 1/((225-x^2)^{3/2)} ?
The integral of 1/((225-x^2)^{3/2)} is x/(225(225-x^2)^{1/2)}+C