What is the integral of 8/(sqrt(36-64x^2)) ?

The integral of 8/(sqrt(36-64x^2)) is arcsin((4x)/3)+C

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integral of 8/(sqrt(36-64x^2))

\int \frac{8}{36 - 64 x ^{2}} d x

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

Solution steps

\int \frac{8}{36 - 64 x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{36 - 64 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{8} d u

= 8 \cdot \int \frac{1}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 8 \cdot \frac{1}{8} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x}{2} d x

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\int \frac{1}{0.37 x - 0.03} d x

What is the integral of 8/(sqrt(36-64x^2)) ?
The integral of 8/(sqrt(36-64x^2)) is arcsin((4x)/3)+C