What is the integral of 1/(sqrt(12x-x^2)) ?

The integral of 1/(sqrt(12x-x^2)) is arcsin(1/6 x-1)+C

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integral of 1/(sqrt(12x-x^2))

\int \frac{1}{12 x - x ^{2}} d x

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

Solution steps

\int \frac{1}{12 x - x ^{2}} d x

Complete the square

12 x - x^{2} : - (x - 6)^{2} + 36

= \int \frac{1}{- ( x - 6 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 36} d u

= \int \frac{1}{- u ^{2} + 36} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

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

Simplify

1 \cdot arcsin (\frac{1}{6} (x - 6)) : arcsin (\frac{1}{6} x - 1)

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{ln ( 5 x )}{5 x})

1. 5^{x}

\int \frac{x}{9 - x ^{2}} d x

\int \frac{1}{sin ^{2} ( 3 x )} d x

\frac{d y}{d x} = \frac{sec ( y )}{( x - 3 ) ^{2}}

What is the integral of 1/(sqrt(12x-x^2)) ?
The integral of 1/(sqrt(12x-x^2)) is arcsin(1/6 x-1)+C