What is the integral of (100)/(w^2sqrt(25-w^2)) ?

The integral of (100)/(w^2sqrt(25-w^2)) is -(4sqrt(-w^2+25))/w+C

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integral of (100)/(w^2sqrt(25-w^2))

\int \frac{100}{w ^{2} 25 - w ^{2}} d w

- \frac{4 - w ^{2} + 25}{w} + C

Solution steps

\int \frac{100}{w ^{2} 25 - w ^{2}} d w

Take the constant out:

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

= 100 \cdot \int \frac{1}{w ^{2} 25 - w ^{2}} d w

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{25 cos ( u ) sin ( u )} d u

= 100 \cdot \int \frac{cot ( u )}{25 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 100 \cdot \frac{1}{25} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= 100 \cdot \frac{1}{25} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 100 \cdot \frac{1}{25} (- cot (u))

Substitute back

u = arcsin (\frac{1}{5} w)

= 100 \cdot \frac{1}{25} (- cot (arcsin (\frac{1}{5} w)))

Simplify

100 \cdot \frac{1}{25} (- cot (arcsin (\frac{1}{5} w))) : - \frac{4 - w ^{2} + 25}{w}

= - \frac{4 - w ^{2} + 25}{w}

Add a constant to the solution

= - \frac{4 - w ^{2} + 25}{w} + C

x y^{^{'}} - 2 y = 3 x^{2}

f (x) = 1 - 6 x, at x = - 1

\frac{9 x}{x}

\int \frac{x ^{2} - 2 x - 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

x \to 1 lim (\frac{3 x - 2}{2 x - 1})

What is the integral of (100)/(w^2sqrt(25-w^2)) ?
The integral of (100)/(w^2sqrt(25-w^2)) is -(4sqrt(-w^2+25))/w+C