What is the integral of-5/(8tsqrt(64t^2-81)) ?

The integral of-5/(8tsqrt(64t^2-81)) is -5/72 arcsec(8/9 t)+C

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integral of-5/(8tsqrt(64t^2-81))

\int - \frac{5}{8 t 64 t ^{2} - 81} d t

- \frac{5}{72} arcsec (\frac{8}{9} t) + C

Solution steps

\int - \frac{5}{8 t 64 t ^{2} - 81} d t

Take the constant out:

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

= - \frac{5}{8} \cdot \int \frac{1}{t 64 t ^{2} - 81} d t

Apply Trigonometric Substitution

: \int \frac{1}{9} d u

= - \frac{5}{8} \cdot \int \frac{1}{9} d u

Take the constant out:

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

= - \frac{5}{8} \cdot \frac{1}{9} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - \frac{5}{8} \cdot \frac{1}{9} \cdot 1 \cdot u

Substitute back

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

= - \frac{5}{8} \cdot \frac{1}{9} \cdot 1 \cdot arcsec (\frac{8}{9} t)

Simplify

- \frac{5}{8} \cdot \frac{1}{9} \cdot 1 \cdot arcsec (\frac{8}{9} t) : - \frac{5}{72} arcsec (\frac{8}{9} t)

= - \frac{5}{72} arcsec (\frac{8}{9} t)

Add a constant to the solution

= - \frac{5}{72} arcsec (\frac{8}{9} t) + C

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What is the integral of-5/(8tsqrt(64t^2-81)) ?
The integral of-5/(8tsqrt(64t^2-81)) is -5/72 arcsec(8/9 t)+C