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

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

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integral of 2/(sqrt(25-2x^2))

\int \frac{2}{25 - 2 x ^{2}} d x

2 arcsin (\frac{2}{5} x) + C

Solution steps

\int \frac{2}{25 - 2 x ^{2}} d x

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

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

Simplify

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

= 2 arcsin (\frac{2}{5} x)

Add a constant to the solution

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

6 x + \frac{4}{x}

n = 1 \sum \infty \frac{1}{n ( ln ( n ) )}

x cos (x) sin (x)

\int - \frac{2}{1 - 2 x} d x

\int_{1}^{4} \frac{x ^{2} + 1}{5 x - x ^{2}} d x

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