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

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

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

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

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

Solution steps

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

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} t)

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

Simplify

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

Add a constant to the solution

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

\frac{4 x}{x ^{2} + 4}

x \to 2 + lim (ln (x - 2))

\int 48 + 2 x - x^{2} d x

\frac{d}{d x} (y + x)

x \to 0 lim (5 x^{3} + 2 x)

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