What is the integral of (3-4x)/(sqrt(16-9x^2)) ?

The integral of (3-4x)/(sqrt(16-9x^2)) is 4/9 sqrt(-9x^2+16)+arcsin(3/4 x)+C

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integral of (3-4x)/(sqrt(16-9x^2))

\int \frac{3 - 4 x}{16 - 9 x ^{2}} d x

\frac{4}{9} - 9 x^{2} + 16 + arcsin (\frac{3}{4} x) + C

Solution steps

\int \frac{3 - 4 x}{16 - 9 x ^{2}} d x

Apply Trigonometric Substitution

= \int \frac{- 16 sin ( u ) + 9}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int - 16 sin (u) + 9 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}{9} (- \int 16 sin (u) d u + \int 9 d u)

\int 16 sin (u) d u = - 16 cos (u)

\int 9 d u = 9 u

= \frac{1}{9} (- (- 16 cos (u)) + 9 u)

Substitute back

u = arcsin (\frac{3}{4} x)

= \frac{1}{9} (- (- 16 cos (arcsin (\frac{3}{4} x))) + 9 arcsin (\frac{3}{4} x))

Simplify

\frac{1}{9} (- (- 16 cos (arcsin (\frac{3}{4} x))) + 9 arcsin (\frac{3}{4} x)) : \frac{4}{9} - 9 x^{2} + 16 + arcsin (\frac{3}{4} x)

= \frac{4}{9} - 9 x^{2} + 16 + arcsin (\frac{3}{4} x)

Add a constant to the solution

= \frac{4}{9} - 9 x^{2} + 16 + arcsin (\frac{3}{4} x) + C

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What is the integral of (3-4x)/(sqrt(16-9x^2)) ?
The integral of (3-4x)/(sqrt(16-9x^2)) is 4/9 sqrt(-9x^2+16)+arcsin(3/4 x)+C