What is the integral of 8/(9tsqrt(81t^2-64)) ?

The integral of 8/(9tsqrt(81t^2-64)) is 1/9 arcsec(9/8 t)+C

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integral of 8/(9tsqrt(81t^2-64))

\int \frac{8}{9 t 81 t ^{2} - 64} d t

\frac{1}{9} arcsec (\frac{9}{8} t) + C

Solution steps

\int \frac{8}{9 t 81 t ^{2} - 64} d t

Take the constant out:

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

= \frac{8}{9} \cdot \int \frac{1}{t 81 t ^{2} - 64} d t

Apply Trigonometric Substitution

= \frac{8}{9} \cdot \int \frac{1}{8} d u

Take the constant out:

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

= \frac{8}{9} \cdot \frac{1}{8} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{8}{9} \cdot \frac{1}{8} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{9}{8} t)

= \frac{8}{9} \cdot \frac{1}{8} \cdot 1 \cdot arcsec (\frac{9}{8} t)

Simplify

\frac{8}{9} \cdot \frac{1}{8} \cdot 1 \cdot arcsec (\frac{9}{8} t) : \frac{1}{9} arcsec (\frac{9}{8} t)

= \frac{1}{9} arcsec (\frac{9}{8} t)

Add a constant to the solution

= \frac{1}{9} arcsec (\frac{9}{8} t) + C

\int (\frac{tan ( x )}{cot ( x )})^{3} d x

\frac{\partial}{\partial x} (y e^{y} - x e^{x})

x \to 3 lim (x^{2} - 5 x + 6)

(y^{3} - 2 x^{2} y) d x + (2 x^{3}) d y = 0

x \to - 6 lim (\frac{( 4 x + 3 ) ^{\frac{2}{3}} - 9}{x ^{2} - 36})

What is the integral of 8/(9tsqrt(81t^2-64)) ?
The integral of 8/(9tsqrt(81t^2-64)) is 1/9 arcsec(9/8 t)+C