What is the integral of sqrt(9-(x-4)^2) ?

The integral of sqrt(9-(x-4)^2) is 9/2 (arcsin(1/3 (x-4))+1/2 sin(2arcsin(1/3 (x-4))))+C

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integral of sqrt(9-(x-4)^2)

\int 9 - (x - 4)^{2} d x

\frac{9}{2} (arcsin (\frac{1}{3} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 4)))) + C

Solution steps

\int 9 - (x - 4)^{2} d x

Apply u-substitution

: \int 9 - u^{2} d u

= \int 9 - u^{2} d u

Apply Trigonometric Substitution

: \int 9 cos^{2} (v) d v

= \int 9 cos^{2} (v) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 9 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

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

= 9 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 9 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 4))))

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 4)))) : \frac{9}{2} (arcsin (\frac{1}{3} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 4))))

= \frac{9}{2} (arcsin (\frac{1}{3} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 4))))

Add a constant to the solution

= \frac{9}{2} (arcsin (\frac{1}{3} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 4)))) + C

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x \to - 5 lim (\frac{x ^{2} - 4}{5 - x})

\int \frac{x ^{2}}{( x ^{2} - 36 ) ^{\frac{5}{2}}} d x

What is the integral of sqrt(9-(x-4)^2) ?
The integral of sqrt(9-(x-4)^2) is 9/2 (arcsin(1/3 (x-4))+1/2 sin(2arcsin(1/3 (x-4))))+C