What is the integral of-2/(sqrt(81-49t^2)) ?

The integral of-2/(sqrt(81-49t^2)) is -2/7 arcsin(7/9 t)+C

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integral of-2/(sqrt(81-49t^2))

\int - \frac{2}{81 - 49 t ^{2}} d t

- \frac{2}{7} arcsin (\frac{7}{9} t) + C

Solution steps

\int - \frac{2}{81 - 49 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}{81 - 49 t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{1}{7} d u

= - 2 \cdot \int \frac{1}{7} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{7} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - 2 \cdot \frac{1}{7} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{7}{9} t)

= - 2 \cdot \frac{1}{7} \cdot 1 \cdot arcsin (\frac{7}{9} t)

Simplify

- 2 \cdot \frac{1}{7} \cdot 1 \cdot arcsin (\frac{7}{9} t) : - \frac{2}{7} arcsin (\frac{7}{9} t)

= - \frac{2}{7} arcsin (\frac{7}{9} t)

Add a constant to the solution

= - \frac{2}{7} arcsin (\frac{7}{9} t) + C

\frac{d}{d x} (\frac{x ^{6}}{x ^{2}})

\int \frac{x ^{2} - 9}{x} d x

y = \frac{1}{x ^{2} + 1}

f (x) = 9 t^{2} + t^{3}

\int 4 sin^{2} (x) cos (x) d x

What is the integral of-2/(sqrt(81-49t^2)) ?
The integral of-2/(sqrt(81-49t^2)) is -2/7 arcsin(7/9 t)+C