What is the integral of (4-32x)/(sqrt(25-16x^2)) ?

The integral of (4-32x)/(sqrt(25-16x^2)) is 2sqrt(-16x^2+25)+arcsin(4/5 x)+C

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integral of (4-32x)/(sqrt(25-16x^2))

\int \frac{4 - 32 x}{25 - 16 x ^{2}} d x

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

Solution steps

\int \frac{4 - 32 x}{25 - 16 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 10 sin (u) + 1 d u

= \int - 10 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 10 sin (u) d u + \int 1 d u

\int 10 sin (u) d u = - 10 cos (u)

\int 1 d u = u

= - (- 10 cos (u)) + u

Substitute back

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

= - (- 10 cos (arcsin (\frac{4}{5} x))) + arcsin (\frac{4}{5} x)

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (8 + 4 x)

\int x x x d x

f (x) = 8 (x - \frac{1}{x})^{3}, at x = 2

n = 0 \sum \infty (3 x)^{n}

y = cot (x), at x = \frac{π}{4}

What is the integral of (4-32x)/(sqrt(25-16x^2)) ?
The integral of (4-32x)/(sqrt(25-16x^2)) is 2sqrt(-16x^2+25)+arcsin(4/5 x)+C