What is the integral of 1/(sqrt(25-x^2)) ?

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

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integral of 1/(sqrt(25-x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

f (x) = 9 x^{2} + 6 x - 6

7 π x^{2}

x \to 0 + lim (12 e^{- \frac{2}{x}})

f (R) = R^{3}

\int (1 + \frac{1}{x})^{- 3} \frac{1}{x ^{2}} d x

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