What is the integral of (5-50x)/(sqrt(16-25x^2)) ?

The integral of (5-50x)/(sqrt(16-25x^2)) is 2sqrt(-25x^2+16)+arcsin(5/4 x)+C

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integral of (5-50x)/(sqrt(16-25x^2))

\int \frac{5 - 50 x}{16 - 25 x ^{2}} d x

2 - 25 x^{2} + 16 + arcsin (\frac{5}{4} x) + C

Solution steps

\int \frac{5 - 50 x}{16 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 8 sin (u) + 1 d u

= \int - 8 sin (u) + 1 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

= - \int 8 sin (u) d u + \int 1 d u

\int 8 sin (u) d u = - 8 cos (u)

\int 1 d u = u

= - (- 8 cos (u)) + u

Substitute back

u = arcsin (\frac{5}{4} x)

= - (- 8 cos (arcsin (\frac{5}{4} x))) + arcsin (\frac{5}{4} x)

Simplify

- (- 8 cos (arcsin (\frac{5}{4} x))) + arcsin (\frac{5}{4} x) : 2 - 25 x^{2} + 16 + arcsin (\frac{5}{4} x)

= 2 - 25 x^{2} + 16 + arcsin (\frac{5}{4} x)

Add a constant to the solution

= 2 - 25 x^{2} + 16 + arcsin (\frac{5}{4} x) + C

θ \to 0 lim (\frac{1 - cos ( θ )}{θ})

\frac{d}{d t} (9 sin (t) - sin (9 t))

x \to 0 lim (\frac{sin ^{8} ( x )}{x})

\int \frac{x - 3}{( x - 5 ) ^{2}} d x

\int_{0}^{x^{2}} e^{- t^{2}} d t

What is the integral of (5-50x)/(sqrt(16-25x^2)) ?
The integral of (5-50x)/(sqrt(16-25x^2)) is 2sqrt(-25x^2+16)+arcsin(5/4 x)+C