What is the integral of 6/(sqrt(25-x^2)) ?

The integral of 6/(sqrt(25-x^2)) is 6arcsin(1/5 x)+C

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integral of 6/(sqrt(25-x^2))

\int \frac{6}{25 - x ^{2}} d x

6 arcsin (\frac{1}{5} x) + C

Solution steps

\int \frac{6}{25 - x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 6 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{1}{5} x)

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

Simplify

= 6 arcsin (\frac{1}{5} x)

Add a constant to the solution

= 6 arcsin (\frac{1}{5} x) + C

t^{2} y^{^{''}} - 2 t y^{^{'}} + 2 y = 3 t^{2} - t^{3}

\int_{2}^{\infty} e^{- 7 p} d p

x \to - 2 lim (\frac{4}{x + 2})

\frac{d}{d x} (3 e^{- \frac{x}{4}})

What is the integral of 6/(sqrt(25-x^2)) ?
The integral of 6/(sqrt(25-x^2)) is 6arcsin(1/5 x)+C