What is the integral of 9/(sqrt(36-81x^2)) ?

The integral of 9/(sqrt(36-81x^2)) is arcsin((3x)/2)+C

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integral of 9/(sqrt(36-81x^2))

\int \frac{9}{36 - 81 x ^{2}} d x

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

Solution steps

\int \frac{9}{36 - 81 x ^{2}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{36 - 81 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{9} d u

= 9 \cdot \int \frac{1}{9} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 9 \cdot \frac{1}{9} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{9}{6} x)

= 9 \cdot \frac{1}{9} \cdot 1 \cdot arcsin (\frac{9}{6} x)

Simplify

9 \cdot \frac{1}{9} \cdot 1 \cdot arcsin (\frac{9}{6} x) : arcsin (\frac{3 x}{2})

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

Add a constant to the solution

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

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What is the integral of 9/(sqrt(36-81x^2)) ?
The integral of 9/(sqrt(36-81x^2)) is arcsin((3x)/2)+C