What is the integral of (50)/(w^2sqrt(25-w^2)) ?

The integral of (50)/(w^2sqrt(25-w^2)) is -(2sqrt(-w^2+25))/w+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (50)/(w^2sqrt(25-w^2))

\int \frac{50}{w ^{2} 25 - w ^{2}} d w

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

Solution steps

\int \frac{50}{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

= 50 \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

= 50 \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

= 50 \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)

= 50 \cdot \frac{1}{25} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 50 \cdot \frac{1}{25} (- cot (u))

Substitute back

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

= 50 \cdot \frac{1}{25} (- cot (arcsin (\frac{1}{5} w)))

Simplify

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

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

Add a constant to the solution

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

\int 4 x + 1 d x

\int θ tan^{2} (θ) d θ

2^{x} + x^{2} e^{x^{2}}

\int 64 e^{- 8 x} sin (8 x) d x

\frac{d}{d x} (ln (x) + x)

What is the integral of (50)/(w^2sqrt(25-w^2)) ?
The integral of (50)/(w^2sqrt(25-w^2)) is -(2sqrt(-w^2+25))/w+C