What is the integral of 1/(sqrt((x-4)(8-x))) ?

The integral of 1/(sqrt((x-4)(8-x))) is arcsin(1/2 (x-6))+C

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integral of 1/(sqrt((x-4)(8-x)))

\int \frac{1}{( x - 4 ) ( 8 - x )} d x

arcsin (\frac{1}{2} (x - 6)) + C

Solution steps

\int \frac{1}{( x - 4 ) ( 8 - x )} d x

\frac{1}{( x - 4 ) ( 8 - x )} = \frac{1 - ( x - 4 ) ( 8 - x )}{( x - 4 ) ( 8 - x )} + 1

= \int \frac{1 - ( x - 4 ) ( 8 - x )}{( x - 4 ) ( 8 - x )} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1 - ( x - 4 ) ( 8 - x )}{( x - 4 ) ( 8 - x )} d x + \int 1 d x

\int \frac{1 - ( x - 4 ) ( 8 - x )}{( x - 4 ) ( 8 - x )} d x = arcsin (\frac{1}{2} (x - 6)) - x

\int 1 d x = x

= arcsin (\frac{1}{2} (x - 6)) - x + x

Simplify

= arcsin (\frac{1}{2} (x - 6))

Add a constant to the solution

= arcsin (\frac{1}{2} (x - 6)) + C

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What is the integral of 1/(sqrt((x-4)(8-x))) ?
The integral of 1/(sqrt((x-4)(8-x))) is arcsin(1/2 (x-6))+C