What is the integral of 1/(sqrt(36-6t^2)) ?

The integral of 1/(sqrt(36-6t^2)) is 1/(sqrt(6))arcsin(t/(sqrt(6)))+C

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integral of 1/(sqrt(36-6t^2))

\int \frac{1}{36 - 6 t ^{2}} d t

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

Solution steps

\int \frac{1}{36 - 6 t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{1}{6} d u

= \int \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{6}{6} t)

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

Simplify

\frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{6} t) : \frac{1}{6} arcsin (\frac{t}{6})

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

Add a constant to the solution

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

f (x) = 2 - x^{2}

{\frac{1}{S ^{3}}}

\int I n (x) d x

ln (1 - 2 x)

y = x + 4

What is the integral of 1/(sqrt(36-6t^2)) ?
The integral of 1/(sqrt(36-6t^2)) is 1/(sqrt(6))arcsin(t/(sqrt(6)))+C