What is the integral of 1/(sqrt(169-x^2)) ?

The integral of 1/(sqrt(169-x^2)) is arcsin(1/13 x)+C

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integral of 1/(sqrt(169-x^2))

\int \frac{1}{169 - x ^{2}} d x

arcsin (\frac{1}{13} x) + C

Solution steps

\int \frac{1}{169 - 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}{13} x)

= 1 \cdot arcsin (\frac{1}{13} x)

Simplify

= arcsin (\frac{1}{13} x)

Add a constant to the solution

= arcsin (\frac{1}{13} x) + C

\int cos (x) ln (2) 2^{x} + 2^{x} sin (x) d x

\frac{d}{d x} ((3 x - 2)^{2} (5 x + 4))

\int (8 x + 3) (7 - x) d x

x \to - \infty lim (- 3 x + x)

h = (t + 1)^{\frac{2}{3}} (2 t^{2} - 1)^{3}

What is the integral of 1/(sqrt(169-x^2)) ?
The integral of 1/(sqrt(169-x^2)) is arcsin(1/13 x)+C