What is the integral of (64)/(w^2sqrt(16-w^2)) ?

The integral of (64)/(w^2sqrt(16-w^2)) is -(4sqrt(-w^2+16))/w+C

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integral of (64)/(w^2sqrt(16-w^2))

\int \frac{64}{w ^{2} 16 - w ^{2}} d w

- \frac{4 - w ^{2} + 16}{w} + C

Solution steps

\int \frac{64}{w ^{2} 16 - w ^{2}} d w

Take the constant out:

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

= 64 \cdot \int \frac{1}{w ^{2} 16 - w ^{2}} d w

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

= 64 \cdot \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 64 \cdot \frac{1}{16} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= 64 \cdot \frac{1}{16} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 64 \cdot \frac{1}{16} (- cot (u))

Substitute back

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

= 64 \cdot \frac{1}{16} (- cot (arcsin (\frac{1}{4} w)))

Simplify

64 \cdot \frac{1}{16} (- cot (arcsin (\frac{1}{4} w))) : - \frac{4 - w ^{2} + 16}{w}

= - \frac{4 - w ^{2} + 16}{w}

Add a constant to the solution

= - \frac{4 - w ^{2} + 16}{w} + C

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What is the integral of (64)/(w^2sqrt(16-w^2)) ?
The integral of (64)/(w^2sqrt(16-w^2)) is -(4sqrt(-w^2+16))/w+C