What is the integral of 1/(sqrt(5-4y^2)) ?

The integral of 1/(sqrt(5-4y^2)) is 1/2 arcsin(2/(sqrt(5))y)+C

Popular >

integral of 1/(sqrt(5-4y^2))

\int \frac{1}{5 - 4 y ^{2}} d y

\frac{1}{2} arcsin (\frac{2}{5} y) + C

Solution steps

\int \frac{1}{5 - 4 y ^{2}} d y

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{2}{5} y) + C

\int \frac{1 + x ^{2}}{x 1 + x ^{2}} d x

\int \frac{2 x}{( x - 2 ) ^{2}} d x

\int \frac{1 - 2 ln ( x )}{x} d x

\int \frac{- 6 x ^{6} - 4 x}{x ^{2}} - 4 e^{6 x} d x

\int 9 x^{6} d x

What is the integral of 1/(sqrt(5-4y^2)) ?
The integral of 1/(sqrt(5-4y^2)) is 1/2 arcsin(2/(sqrt(5))y)+C