What is the integral of 6/(sqrt(25-16t^2)) ?

The integral of 6/(sqrt(25-16t^2)) is 3/2 arcsin(4/5 t)+C

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integral of 6/(sqrt(25-16t^2))

\int \frac{6}{25 - 16 t ^{2}} d t

\frac{3}{2} arcsin (\frac{4}{5} t) + C

Solution steps

\int \frac{6}{25 - 16 t ^{2}} d t

Take the constant out:

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

= 6 \cdot \int \frac{1}{25 - 16 t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

u = arcsin (\frac{4}{5} t)

= 6 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{5} t)

Simplify

6 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{5} t) : \frac{3}{2} arcsin (\frac{4}{5} t)

= \frac{3}{2} arcsin (\frac{4}{5} t)

Add a constant to the solution

= \frac{3}{2} arcsin (\frac{4}{5} t) + C

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What is the integral of 6/(sqrt(25-16t^2)) ?
The integral of 6/(sqrt(25-16t^2)) is 3/2 arcsin(4/5 t)+C