What is the integral of 1/(sqrt(25-16x^2)) ?

The integral of 1/(sqrt(25-16x^2)) is 1/4 arcsin(4/5 x)+C

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integral of 1/(sqrt(25-16x^2))

\int \frac{1}{25 - 16 x ^{2}} d x

\frac{1}{4} arcsin (\frac{4}{5} x) + C

Solution steps

\int \frac{1}{25 - 16 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

= \int \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot 1 \cdot u

Substitute back

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

= \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{5} x)

Simplify

= \frac{1}{4} arcsin (\frac{4}{5} x)

Add a constant to the solution

= \frac{1}{4} arcsin (\frac{4}{5} x) + C

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

f (x) = 3 x - 2 x, (1, 1)

z \to + 10 lim (\frac{1}{z - 10})

- 5 csc (x) + sec (x)

2 ln (3 x)

What is the integral of 1/(sqrt(25-16x^2)) ?
The integral of 1/(sqrt(25-16x^2)) is 1/4 arcsin(4/5 x)+C