What is the integral of 1/((49-x^2)^{3/2)} ?

The integral of 1/((49-x^2)^{3/2)} is x/(49(49-x^2)^{1/2)}+C

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integral of 1/((49-x^2)^{3/2)}

\int \frac{1}{( 49 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{49 ( 49 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 49 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{49 cos ^{2} ( u )} d u

= \int \frac{1}{49 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}{49} \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}{49} tan (u)

Substitute back

u = arcsin (\frac{1}{7} x)

= \frac{1}{49} tan (arcsin (\frac{1}{7} x))

Simplify

\frac{1}{49} tan (arcsin (\frac{1}{7} x)) : \frac{x}{49 ( 49 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{49 ( 49 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{49 ( 49 - x ^{2} ) ^{\frac{1}{2}}} + C

f (x) = arccos (2 x)

(- 10, - 7), (- 5, - 9)

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

\frac{\partial}{\partial s} (1 + s t)

What is the integral of 1/((49-x^2)^{3/2)} ?
The integral of 1/((49-x^2)^{3/2)} is x/(49(49-x^2)^{1/2)}+C