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

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

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

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

arcsin (\frac{x}{a}) + C

Solution steps

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

a - x^{2} = a (1 - \frac{x ^{2}}{a})

= \int \frac{1}{a ( 1 - \frac{x ^{2}}{a} )} d x

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{a} = (\frac{x}{a})^{2}

= \int \frac{1}{a ( 1 - ( \frac{x}{a} ) ^{2} )} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

a (1 - (\frac{x}{a})^{2}) = a (1 - (\frac{x}{a})^{2})

= \int \frac{1}{a 1 - ( \frac{x}{a} ) ^{2}} d x

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{1}{1 - ( \frac{x}{a} ) ^{2}} d x

Apply u-substitution

: \int \frac{a}{1 - u ^{2}} d u

= \frac{1}{a} \cdot \int \frac{a}{1 - u ^{2}} d u

\frac{a}{1 - u ^{2}} = a \frac{1}{1 - u ^{2}}

= \frac{1}{a} \cdot \int a \frac{1}{1 - u ^{2}} d u

Take the constant out:

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

= \frac{1}{a} a \cdot \int \frac{1}{1 - u ^{2}} d u

Simplify

\frac{1}{1 - u ^{2}} : \frac{1}{1 - u ^{2}}

= \frac{1}{a} a \cdot \int \frac{1}{1 - u ^{2}} d u

Simplify

= \frac{1}{a} a \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = arcsin (u)

= \frac{1}{a} a arcsin (u)

Substitute back

u = \frac{x}{a}

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

Simplify

\frac{1}{a} a arcsin (\frac{x}{a}) : arcsin (\frac{x}{a})

= arcsin (\frac{x}{a})

Add a constant to the solution

= arcsin (\frac{x}{a}) + C

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