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

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

Popular >

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

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

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

Solution steps

\int \frac{1}{9 - 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}{3} x)

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (3 x^{4} - 2 x^{3} + 1)

\int 5 csc^{2} (x) d x

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

\int e^{(\frac{x ^{2}}{2})} d x

x \to 7 + lim (3)

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