What is the integral of (196)/(w^2sqrt(49-w^2)) ?

The integral of (196)/(w^2sqrt(49-w^2)) is -(4sqrt(-w^2+49))/w+C

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integral of (196)/(w^2sqrt(49-w^2))

\int \frac{196}{w ^{2} 49 - w ^{2}} d w

- \frac{4 - w ^{2} + 49}{w} + C

Solution steps

\int \frac{196}{w ^{2} 49 - w ^{2}} d w

Take the constant out:

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

= 196 \cdot \int \frac{1}{w ^{2} 49 - w ^{2}} d w

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{49 cos ( u ) sin ( u )} d u

= 196 \cdot \int \frac{cot ( u )}{49 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 196 \cdot \frac{1}{49} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= 196 \cdot \frac{1}{49} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 196 \cdot \frac{1}{49} (- cot (u))

Substitute back

u = arcsin (\frac{1}{7} w)

= 196 \cdot \frac{1}{49} (- cot (arcsin (\frac{1}{7} w)))

Simplify

196 \cdot \frac{1}{49} (- cot (arcsin (\frac{1}{7} w))) : - \frac{4 - w ^{2} + 49}{w}

= - \frac{4 - w ^{2} + 49}{w}

Add a constant to the solution

= - \frac{4 - w ^{2} + 49}{w} + C

\int \frac{1}{2 x + 3} d x

\frac{\partial}{\partial y} (2 x (5 + y)^{- 1})

\frac{sec ( 9 x ^{2} )}{tan ( 2 x ) \cdot cos ( x )}

y^{^{'}} = 3

1 + 6 x^{\frac{3}{2}}

What is the integral of (196)/(w^2sqrt(49-w^2)) ?
The integral of (196)/(w^2sqrt(49-w^2)) is -(4sqrt(-w^2+49))/w+C