What is the integral of 1/(sqrt(36-x^2)) ?

The integral of 1/(sqrt(36-x^2)) is arcsin(1/6 x)+C

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integral of 1/(sqrt(36-x^2))

\int \frac{1}{36 - x ^{2}} d x

arcsin (\frac{1}{6} x) + C

Solution steps

\int \frac{1}{36 - 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}{6} x)

= 1 \cdot arcsin (\frac{1}{6} x)

Simplify

= arcsin (\frac{1}{6} x)

Add a constant to the solution

= arcsin (\frac{1}{6} x) + C

\frac{1}{1 - x ^{3}}, 0

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

\frac{d}{d x} (\frac{5}{3} x^{3} + x^{2} - 1)

\int \frac{sin ^{7} ( ln ( x ) )}{x} d x

\frac{1}{x ^{8}} y^{^{'}} = 12 + 15 y

What is the integral of 1/(sqrt(36-x^2)) ?
The integral of 1/(sqrt(36-x^2)) is arcsin(1/6 x)+C