What is the integral of 8/(w^2sqrt(4-w^2)) ?

The integral of 8/(w^2sqrt(4-w^2)) is -(2sqrt(-w^2+4))/w+C

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integral of 8/(w^2sqrt(4-w^2))

\int \frac{8}{w ^{2} 4 - w ^{2}} d w

- \frac{2 - w ^{2} + 4}{w} + C

Solution steps

\int \frac{8}{w ^{2} 4 - w ^{2}} d w

Take the constant out:

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

= 8 \cdot \int \frac{1}{w ^{2} 4 - w ^{2}} d w

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{4 cos ( u ) sin ( u )} d u

= 8 \cdot \int \frac{cot ( u )}{4 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= 8 \cdot \frac{1}{4} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= 8 \cdot \frac{1}{4} (- cot (u))

Substitute back

u = arcsin (\frac{1}{2} w)

= 8 \cdot \frac{1}{4} (- cot (arcsin (\frac{1}{2} w)))

Simplify

8 \cdot \frac{1}{4} (- cot (arcsin (\frac{1}{2} w))) : - \frac{2 - w ^{2} + 4}{w}

= - \frac{2 - w ^{2} + 4}{w}

Add a constant to the solution

= - \frac{2 - w ^{2} + 4}{w} + C

\frac{d}{d x} (x^{4} - 3)

\frac{d y}{d x} = e^{- a \cdot x} - b \cdot y, y (0) = 0

\int \frac{( 2 r + 1 )}{r ^{2} - 3 r + 2} d r

x \to 10 lim (\frac{1}{( x - 10 ) ^{2}})

\frac{e ^{- 4 s}}{s ^{3}}

What is the integral of 8/(w^2sqrt(4-w^2)) ?
The integral of 8/(w^2sqrt(4-w^2)) is -(2sqrt(-w^2+4))/w+C