What is the integral of 1/((64-x^2)^{3/2)} ?

The integral of 1/((64-x^2)^{3/2)} is x/(64(64-x^2)^{1/2)}+C

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integral of 1/((64-x^2)^{3/2)}

\int \frac{1}{( 64 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{64 ( 64 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 64 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{64 cos ^{2} ( u )} d u

= \int \frac{1}{64 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}{64} \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}{64} tan (u)

Substitute back

u = arcsin (\frac{1}{8} x)

= \frac{1}{64} tan (arcsin (\frac{1}{8} x))

Simplify

\frac{1}{64} tan (arcsin (\frac{1}{8} x)) : \frac{x}{64 ( 64 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{64 ( 64 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{64 ( 64 - x ^{2} ) ^{\frac{1}{2}}} + C

\frac{220 x}{( x + 2 ) ^{2}}

\int (1 - \frac{3}{y} + x) d y

\int_{1}^{4} (x^{3} + \frac{1}{4 x ^{3}}) d x

n = 0 \sum \infty \frac{n ^{6}}{2 ^{n}}

n = 1 \sum \infty - 1^{n}

What is the integral of 1/((64-x^2)^{3/2)} ?
The integral of 1/((64-x^2)^{3/2)} is x/(64(64-x^2)^{1/2)}+C