What is the integral of 2/(sqrt(36-25y^2)) ?

The integral of 2/(sqrt(36-25y^2)) is 2/5 arcsin(5/6 y)+C

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integral of 2/(sqrt(36-25y^2))

\int \frac{2}{36 - 25 y ^{2}} d y

\frac{2}{5} arcsin (\frac{5}{6} y) + C

Solution steps

\int \frac{2}{36 - 25 y ^{2}} d y

Take the constant out:

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

= 2 \cdot \int \frac{1}{36 - 25 y ^{2}} d y

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= 2 \cdot \int \frac{1}{5} d u

Take the constant out:

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

= 2 \cdot \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{5} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{5}{6} y)

= 2 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{6} y)

Simplify

2 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{6} y) : \frac{2}{5} arcsin (\frac{5}{6} y)

= \frac{2}{5} arcsin (\frac{5}{6} y)

Add a constant to the solution

= \frac{2}{5} arcsin (\frac{5}{6} y) + C

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What is the integral of 2/(sqrt(36-25y^2)) ?
The integral of 2/(sqrt(36-25y^2)) is 2/5 arcsin(5/6 y)+C