What is the integral of 1/(sqrt((25-y^2)-x^2)) ?

The integral of 1/(sqrt((25-y^2)-x^2)) is arcsin(x/(sqrt(25-y^2)))+C

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integral of 1/(sqrt((25-y^2)-x^2))

\int \frac{1}{( 25 - y ^{2} ) - x ^{2}} d x

arcsin (\frac{x}{25 - y ^{2}}) + C

Solution steps

\int \frac{1}{25 - y ^{2} - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = arcsin (\frac{1}{25 - y ^{2}} x)

= 1 \cdot arcsin (\frac{1}{25 - y ^{2}} x)

Simplify

1 \cdot arcsin (\frac{1}{25 - y ^{2}} x) : arcsin (\frac{x}{25 - y ^{2}})

= arcsin (\frac{x}{25 - y ^{2}})

Add a constant to the solution

= arcsin (\frac{x}{25 - y ^{2}}) + C

sin (2 t) + 2 cos (2 t) + e^{- t} sin (2 t)

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

n = 1 \sum \infty 8^{n} x^{n} n!

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

x \cdot \frac{d y}{d x} = x tan (\frac{y}{x}) + y

What is the integral of 1/(sqrt((25-y^2)-x^2)) ?
The integral of 1/(sqrt((25-y^2)-x^2)) is arcsin(x/(sqrt(25-y^2)))+C