What is the integral of (36)/((1-x^2)^{3/2)} ?

The integral of (36)/((1-x^2)^{3/2)} is (36x)/((1-x^2)^{1/2)}+C

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integral of (36)/((1-x^2)^{3/2)}

\int \frac{36}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{36 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{36}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 36 \cdot \int \frac{1}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 36 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 36 tan (u)

Substitute back

u = arcsin (x)

= 36 tan (arcsin (x))

Simplify

36 tan (arcsin (x)) : \frac{36 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{36 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{36 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

(- 5) (- 5.1)

\frac{d}{d x} (cos^{\frac{7}{4}} (4 x^{3}))

\int_{1}^{8} x + 1 d x

f (x) = x^{\frac{3}{2}} + x + 4

x e^{x}

What is the integral of (36)/((1-x^2)^{3/2)} ?
The integral of (36)/((1-x^2)^{3/2)} is (36x)/((1-x^2)^{1/2)}+C