What is the integral of (8-3x)/(sqrt(25-64x^2)) ?

The integral of (8-3x)/(sqrt(25-64x^2)) is 3/64 sqrt(-64x^2+25)+arcsin(8/5 x)+C

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integral of (8-3x)/(sqrt(25-64x^2))

\int \frac{8 - 3 x}{25 - 64 x ^{2}} d x

\frac{3}{64} - 64 x^{2} + 25 + arcsin (\frac{8}{5} x) + C

Solution steps

\int \frac{8 - 3 x}{25 - 64 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{- 15 sin ( u ) + 64}{64} d u

= \int \frac{- 15 sin ( u ) + 64}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int - 15 sin (u) + 64 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

= \frac{1}{64} (- \int 15 sin (u) d u + \int 64 d u)

\int 15 sin (u) d u = - 15 cos (u)

\int 64 d u = 64 u

= \frac{1}{64} (- (- 15 cos (u)) + 64 u)

Substitute back

u = arcsin (\frac{8}{5} x)

= \frac{1}{64} (- (- 15 cos (arcsin (\frac{8}{5} x))) + 64 arcsin (\frac{8}{5} x))

Simplify

\frac{1}{64} (- (- 15 cos (arcsin (\frac{8}{5} x))) + 64 arcsin (\frac{8}{5} x)) : \frac{3}{64} - 64 x^{2} + 25 + arcsin (\frac{8}{5} x)

= \frac{3}{64} - 64 x^{2} + 25 + arcsin (\frac{8}{5} x)

Add a constant to the solution

= \frac{3}{64} - 64 x^{2} + 25 + arcsin (\frac{8}{5} x) + C

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What is the integral of (8-3x)/(sqrt(25-64x^2)) ?
The integral of (8-3x)/(sqrt(25-64x^2)) is 3/64 sqrt(-64x^2+25)+arcsin(8/5 x)+C