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

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

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integral of 1/(sqrt(r^2-x^2))

\int \frac{1}{r ^{2} - x ^{2}} d x

arcsin (\frac{x}{r}) + C

Solution steps

\int \frac{1}{r ^{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}{r} x)

= 1 \cdot arcsin (\frac{1}{r} x)

Simplify

1 \cdot arcsin (\frac{1}{r} x) : arcsin (\frac{x}{r})

= arcsin (\frac{x}{r})

Add a constant to the solution

= arcsin (\frac{x}{r}) + C

\frac{d y}{d x}, x y = 6

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

y = \frac{7}{x - 1}

4 cos (5 ln (x))

(cos (t) + t sin (t)) + (sin (t) - t cos (t))

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