What is the integral of 3sqrt(4-25s^2) ?

The integral of 3sqrt(4-25s^2) is 3/5 (2arcsin(5/2 s)+sin(2arcsin(5/2 s)))+C

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integral of 3sqrt(4-25s^2)

\int 3 4 - 25 s^{2} d s

\frac{3}{5} (2 arcsin (\frac{5}{2} s) + sin (2 arcsin (\frac{5}{2} s))) + C

Solution steps

\int 3 4 - 25 s^{2} d s

Take the constant out:

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

= 3 \cdot \int 4 - 25 s^{2} d s

Apply Trigonometric Substitution

: \int \frac{4}{5} cos^{2} (u) d u

= 3 \cdot \int \frac{4}{5} cos^{2} (u) d u

Take the constant out:

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

= 3 \cdot \frac{4}{5} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 3 \cdot \frac{4}{5} \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 3 \cdot \frac{4}{5} \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

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

= 3 \cdot \frac{4}{5} \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

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

= 3 \cdot \frac{4}{5} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{5}{2} s)

= 3 \cdot \frac{4}{5} \cdot \frac{1}{2} (arcsin (\frac{5}{2} s) + \frac{1}{2} sin (2 arcsin (\frac{5}{2} s)))

Simplify

3 \cdot \frac{4}{5} \cdot \frac{1}{2} (arcsin (\frac{5}{2} s) + \frac{1}{2} sin (2 arcsin (\frac{5}{2} s))) : \frac{3}{5} (2 arcsin (\frac{5}{2} s) + sin (2 arcsin (\frac{5}{2} s)))

= \frac{3}{5} (2 arcsin (\frac{5}{2} s) + sin (2 arcsin (\frac{5}{2} s)))

Add a constant to the solution

= \frac{3}{5} (2 arcsin (\frac{5}{2} s) + sin (2 arcsin (\frac{5}{2} s))) + C

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What is the integral of 3sqrt(4-25s^2) ?
The integral of 3sqrt(4-25s^2) is 3/5 (2arcsin(5/2 s)+sin(2arcsin(5/2 s)))+C