What is the integral of (72)/(w^2sqrt(36-w^2)) ?

The integral of (72)/(w^2sqrt(36-w^2)) is -(2sqrt(-w^2+36))/w+C

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integral of (72)/(w^2sqrt(36-w^2))

\int \frac{72}{w ^{2} 36 - w ^{2}} d w

- \frac{2 - w ^{2} + 36}{w} + C

Solution steps

\int \frac{72}{w ^{2} 36 - w ^{2}} d w

Take the constant out:

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

= 72 \cdot \int \frac{1}{w ^{2} 36 - w ^{2}} d w

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{36 cos ( u ) sin ( u )} d u

= 72 \cdot \int \frac{cot ( u )}{36 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 72 \cdot \frac{1}{36} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= 72 \cdot \frac{1}{36} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 72 \cdot \frac{1}{36} (- cot (u))

Substitute back

u = arcsin (\frac{1}{6} w)

= 72 \cdot \frac{1}{36} (- cot (arcsin (\frac{1}{6} w)))

Simplify

72 \cdot \frac{1}{36} (- cot (arcsin (\frac{1}{6} w))) : - \frac{2 - w ^{2} + 36}{w}

= - \frac{2 - w ^{2} + 36}{w}

Add a constant to the solution

= - \frac{2 - w ^{2} + 36}{w} + C

y^{^{'}} + sin (t) y = sin (t), y (0) = 8

\frac{\partial}{\partial x} (3 x^{2} y - 4 x y^{2})

\frac{t}{( t - 4 ) ^{2}}

f (x) = x^{2} - x, (3, 6)

a^{\frac{2}{3}}

What is the integral of (72)/(w^2sqrt(36-w^2)) ?
The integral of (72)/(w^2sqrt(36-w^2)) is -(2sqrt(-w^2+36))/w+C