What is the integral of (5-50x)/(sqrt(36-25x^2)) ?

The integral of (5-50x)/(sqrt(36-25x^2)) is 2sqrt(-25x^2+36)+arcsin(5/6 x)+C

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integral of (5-50x)/(sqrt(36-25x^2))

\int \frac{5 - 50 x}{36 - 25 x ^{2}} d x

2 - 25 x^{2} + 36 + arcsin (\frac{5}{6} x) + C

Solution steps

\int \frac{5 - 50 x}{36 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 12 sin (u) + 1 d u

= \int - 12 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 12 sin (u) d u + \int 1 d u

\int 12 sin (u) d u = - 12 cos (u)

\int 1 d u = u

= - (- 12 cos (u)) + u

Substitute back

u = arcsin (\frac{5}{6} x)

= - (- 12 cos (arcsin (\frac{5}{6} x))) + arcsin (\frac{5}{6} x)

Simplify

- (- 12 cos (arcsin (\frac{5}{6} x))) + arcsin (\frac{5}{6} x) : 2 - 25 x^{2} + 36 + arcsin (\frac{5}{6} x)

= 2 - 25 x^{2} + 36 + arcsin (\frac{5}{6} x)

Add a constant to the solution

= 2 - 25 x^{2} + 36 + arcsin (\frac{5}{6} x) + C

\int \frac{x ^{2}}{2} - 2 x d x

\int \frac{1}{( x + 1 ) ( 3 x + 1 )} d x

\int \frac{x}{5 + x - x ^{2}} d x

\int x 2 x + 3 d x

\frac{\partial}{\partial x} (cos^{6} (x^{9} y^{5}))

What is the integral of (5-50x)/(sqrt(36-25x^2)) ?
The integral of (5-50x)/(sqrt(36-25x^2)) is 2sqrt(-25x^2+36)+arcsin(5/6 x)+C