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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= \int \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

\frac{5}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

\int (4 x^{2} + 8 x + 9)^{6} (x + 1) d x

arccos (x)

x \to 0 lim (\frac{e ^{x}}{x ^{2}})

\frac{d ^{4}}{d x ^{4}} (x^{- 2})

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