What is the integral of 1/(sqrt(289-x^2)) ?

The integral of 1/(sqrt(289-x^2)) is arcsin(1/17 x)+C

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integral of 1/(sqrt(289-x^2))

\int \frac{1}{289 - x ^{2}} d x

arcsin (\frac{1}{17} x) + C

Solution steps

\int \frac{1}{289 - 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}{17} x)

= 1 \cdot arcsin (\frac{1}{17} x)

Simplify

= arcsin (\frac{1}{17} x)

Add a constant to the solution

= arcsin (\frac{1}{17} x) + C

\frac{d}{d x} (sin (4 x) - \frac{1}{3} cos^{2} (4 x))

n = 1 \sum \infty 3 (\frac{1}{2})^{2 n - 2}

\int_{- \infty}^{π} e^{x} d x

(x^{2} + 4) y = 8, (2, 1)

4 x, 4 x

What is the integral of 1/(sqrt(289-x^2)) ?
The integral of 1/(sqrt(289-x^2)) is arcsin(1/17 x)+C