What is the integral of 5/(sqrt(k-x^2)) ?

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

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integral of 5/(sqrt(k-x^2))

\int \frac{5}{k - x ^{2}} d x

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

Solution steps

\int \frac{5}{k - x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{k - x ^{2}} d x

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

= 5 \cdot \int \frac{1}{k ( 1 - \frac{x ^{2}}{k} )} d x

Apply exponent rule:

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

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

= 5 \cdot \int \frac{1}{k ( 1 - ( \frac{x}{k} ) ^{2} )} d x

Apply exponent rule:

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

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

= 5 \cdot \int \frac{1}{k 1 - ( \frac{x}{k} ) ^{2}} d x

Take the constant out:

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

= 5 \cdot \frac{1}{k} \cdot \int \frac{1}{1 - ( \frac{x}{k} ) ^{2}} d x

Apply u-substitution

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

= 5 \cdot \frac{1}{k} \cdot \int \frac{k}{1 - u ^{2}} d u

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

= 5 \cdot \frac{1}{k} \cdot \int k \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

= 5 \cdot \frac{1}{k} k \cdot \int \frac{1}{1 - u ^{2}} d u

Simplify

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

= 5 \cdot \frac{1}{k} k \cdot \int \frac{1}{1 - u ^{2}} d u

Simplify

= 5 \cdot \frac{1}{k} k \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

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

= 5 \cdot \frac{1}{k} k arcsin (u)

Substitute back

u = \frac{x}{k}

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

Simplify

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

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

Add a constant to the solution

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

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