What is the integral of-2/(sqrt(100-9t^2)) ?

The integral of-2/(sqrt(100-9t^2)) is -2/3 arcsin(3/10 t)+C

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integral of-2/(sqrt(100-9t^2))

\int - \frac{2}{100 - 9 t ^{2}} d t

- \frac{2}{3} arcsin (\frac{3}{10} t) + C

Solution steps

\int - \frac{2}{100 - 9 t ^{2}} d t

Take the constant out:

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

= - 2 \cdot \int \frac{1}{100 - 9 t ^{2}} d t

Apply Trigonometric Substitution

= - 2 \cdot \int \frac{1}{3} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - 2 \cdot \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{3}{10} t)

= - 2 \cdot \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{10} t)

Simplify

- 2 \cdot \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{10} t) : - \frac{2}{3} arcsin (\frac{3}{10} t)

= - \frac{2}{3} arcsin (\frac{3}{10} t)

Add a constant to the solution

= - \frac{2}{3} arcsin (\frac{3}{10} t) + C

What is the integral of-2/(sqrt(100-9t^2)) ?
The integral of-2/(sqrt(100-9t^2)) is -2/3 arcsin(3/10 t)+C