What is the integral of 1/(sqrt(16-9x^2)) ?

The integral of 1/(sqrt(16-9x^2)) is 1/3 arcsin(3/4 x)+C

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integral of 1/(sqrt(16-9x^2))

\int \frac{1}{16 - 9 x ^{2}} d x

\frac{1}{3} arcsin (\frac{3}{4} x) + C

Solution steps

\int \frac{1}{16 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= \int \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{3}{4} x)

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{4} x)

Simplify

= \frac{1}{3} arcsin (\frac{3}{4} x)

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{3}{4} x) + C

\frac{\partial}{\partial x} (a^{- x})

\int_{0}^{π} 8 sin (\frac{x}{2}) d x

y^{^{''}} - 6 y^{^{'}} + 8 y = - 2 cos (4 t)

\int_{0}^{2} 8 t d t

What is the integral of 1/(sqrt(16-9x^2)) ?
The integral of 1/(sqrt(16-9x^2)) is 1/3 arcsin(3/4 x)+C