What is the integral of-7/(4tsqrt(16t^2-36)) ?

The integral of-7/(4tsqrt(16t^2-36)) is -7/24 arcsec((2t)/3)+C

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integral of-7/(4tsqrt(16t^2-36))

\int - \frac{7}{4 t 16 t ^{2} - 36} d t

- \frac{7}{24} arcsec (\frac{2 t}{3}) + C

Solution steps

\int - \frac{7}{4 t 16 t ^{2} - 36} d t

Take the constant out:

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

= - \frac{7}{4} \cdot \int \frac{1}{t 16 t ^{2} - 36} d t

Apply Trigonometric Substitution

= - \frac{7}{4} \cdot \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{7}{4} \cdot \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - \frac{7}{4} \cdot \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{4}{6} t)

= - \frac{7}{4} \cdot \frac{1}{6} \cdot 1 \cdot arcsec (\frac{4}{6} t)

Simplify

- \frac{7}{4} \cdot \frac{1}{6} \cdot 1 \cdot arcsec (\frac{4}{6} t) : - \frac{7}{24} arcsec (\frac{2 t}{3})

= - \frac{7}{24} arcsec (\frac{2 t}{3})

Add a constant to the solution

= - \frac{7}{24} arcsec (\frac{2 t}{3}) + C

What is the integral of-7/(4tsqrt(16t^2-36)) ?
The integral of-7/(4tsqrt(16t^2-36)) is -7/24 arcsec((2t)/3)+C