What is the integral of 4/(sqrt(64-36x^2)) ?

The integral of 4/(sqrt(64-36x^2)) is 2/3 arcsin(3/4 x)+C

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integral of 4/(sqrt(64-36x^2))

\int \frac{4}{64 - 36 x ^{2}} d x

\frac{2}{3} arcsin (\frac{3}{4} x) + C

Solution steps

\int \frac{4}{64 - 36 x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{64 - 36 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} d u

= 4 \cdot \int \frac{1}{6} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{6}{8} x)

= 4 \cdot \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{8} x)

Simplify

4 \cdot \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{8} x) : \frac{2}{3} arcsin (\frac{3}{4} x)

= \frac{2}{3} arcsin (\frac{3}{4} x)

Add a constant to the solution

= \frac{2}{3} arcsin (\frac{3}{4} x) + C

\frac{d}{d x} (3 π)

x \to 2 - lim (9 - x^{2})

\frac{d}{d y} (- y + 1)

\frac{d}{d x} (- \frac{12}{x ^{6}})

\int \frac{x + 1}{x ^{3}} d x

What is the integral of 4/(sqrt(64-36x^2)) ?
The integral of 4/(sqrt(64-36x^2)) is 2/3 arcsin(3/4 x)+C