What is the integral of 1/((16-x^2)^{3/2)} ?

The integral of 1/((16-x^2)^{3/2)} is x/(16(16-x^2)^{1/2)}+C

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integral of 1/((16-x^2)^{3/2)}

\int \frac{1}{( 16 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{16 ( 16 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 16 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 cos ^{2} ( u )} d u

= \int \frac{1}{16 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}{16} \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}{16} tan (u)

Substitute back

u = arcsin (\frac{1}{4} x)

= \frac{1}{16} tan (arcsin (\frac{1}{4} x))

Simplify

\frac{1}{16} tan (arcsin (\frac{1}{4} x)) : \frac{x}{16 ( 16 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{16 ( 16 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{16 ( 16 - x ^{2} ) ^{\frac{1}{2}}} + C

\int e^{\frac{x}{n}} d x

f (x) = \frac{1}{( 3 x ^{2} - 1 ) ^{2}}

f (x) = x^{2} + 3, (3, 12)

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

f (x) = 3 csc (4 x)

What is the integral of 1/((16-x^2)^{3/2)} ?
The integral of 1/((16-x^2)^{3/2)} is x/(16(16-x^2)^{1/2)}+C