What is the integral of (16)/(sqrt(49-36x^2)) ?

The integral of (16)/(sqrt(49-36x^2)) is 8/3 arcsin(6/7 x)+C

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integral of (16)/(sqrt(49-36x^2))

\int \frac{16}{49 - 36 x ^{2}} d x

\frac{8}{3} arcsin (\frac{6}{7} x) + C

Solution steps

\int \frac{16}{49 - 36 x ^{2}} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{49 - 36 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} d u

= 16 \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

= 16 \cdot \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 16 \cdot \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{6}{7} x)

= 16 \cdot \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{7} x)

Simplify

16 \cdot \frac{1}{6} \cdot 1 \cdot arcsin (\frac{6}{7} x) : \frac{8}{3} arcsin (\frac{6}{7} x)

= \frac{8}{3} arcsin (\frac{6}{7} x)

Add a constant to the solution

= \frac{8}{3} arcsin (\frac{6}{7} x) + C

\frac{d}{d s} (\frac{s}{t})

\int \frac{3}{6 x - x ^{2}} d x

\frac{d}{d x} (x^{3} cos (x) - 9 x sin (x) - 9 cos (x))

x \to 0 lim (\frac{0}{x ^{4}})

- 3 x^{\frac{3}{2}} - 1

What is the integral of (16)/(sqrt(49-36x^2)) ?
The integral of (16)/(sqrt(49-36x^2)) is 8/3 arcsin(6/7 x)+C