What is the integral of 1/(sqrt(6-x^2)) ?

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

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integral of 1/(sqrt(6-x^2))

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

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

Solution steps

\int \frac{1}{6 - 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}{6} x)

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

Simplify

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

Add a constant to the solution

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

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\int_{0}^{2} (8 t i - t^{3} j + 3 t^{5} k) d t

\int \frac{e ^{\frac{1}{x ^{9}}}}{x ^{10}} d x

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