What is the integral of (7-5x)/(sqrt(64-49x^2)) ?

The integral of (7-5x)/(sqrt(64-49x^2)) is 5/49 sqrt(-49x^2+64)+arcsin(7/8 x)+C

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integral of (7-5x)/(sqrt(64-49x^2))

\int \frac{7 - 5 x}{64 - 49 x ^{2}} d x

\frac{5}{49} - 49 x^{2} + 64 + arcsin (\frac{7}{8} x) + C

Solution steps

\int \frac{7 - 5 x}{64 - 49 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{- 40 sin ( u ) + 49}{49} d u

= \int \frac{- 40 sin ( u ) + 49}{49} d u

Take the constant out:

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

= \frac{1}{49} \cdot \int - 40 sin (u) + 49 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}{49} (- \int 40 sin (u) d u + \int 49 d u)

\int 40 sin (u) d u = - 40 cos (u)

\int 49 d u = 49 u

= \frac{1}{49} (- (- 40 cos (u)) + 49 u)

Substitute back

u = arcsin (\frac{7}{8} x)

= \frac{1}{49} (- (- 40 cos (arcsin (\frac{7}{8} x))) + 49 arcsin (\frac{7}{8} x))

Simplify

\frac{1}{49} (- (- 40 cos (arcsin (\frac{7}{8} x))) + 49 arcsin (\frac{7}{8} x)) : \frac{5}{49} - 49 x^{2} + 64 + arcsin (\frac{7}{8} x)

= \frac{5}{49} - 49 x^{2} + 64 + arcsin (\frac{7}{8} x)

Add a constant to the solution

= \frac{5}{49} - 49 x^{2} + 64 + arcsin (\frac{7}{8} x) + C

\int \frac{( 2 x - 3 )}{x ^{2} + 6 x + 4} d x

f (x) = e^{x} + e^{- x}

t \to 0 lim (\frac{1 - cos ( t )}{t})

\frac{d}{d x} (4 cos (\frac{x}{2}))

y^{^{''}} - 3 y = 0

What is the integral of (7-5x)/(sqrt(64-49x^2)) ?
The integral of (7-5x)/(sqrt(64-49x^2)) is 5/49 sqrt(-49x^2+64)+arcsin(7/8 x)+C