What is the integral of 1/(sqrt(16-49x^2)) ?

The integral of 1/(sqrt(16-49x^2)) is 1/7 arcsin(7/4 x)+C

Popular >

integral of 1/(sqrt(16-49x^2))

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

\frac{1}{7} arcsin (\frac{7}{4} x) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{7} d u

= \int \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{7} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{7}{4} x)

= \frac{1}{7} \cdot 1 \cdot arcsin (\frac{7}{4} x)

Simplify

= \frac{1}{7} arcsin (\frac{7}{4} x)

Add a constant to the solution

= \frac{1}{7} arcsin (\frac{7}{4} x) + C

\frac{d}{d x} (\frac{x ^{2} + 2 x + 3}{x ^{2}})

\int \frac{1}{( y + 2 ) ^{2}} d y

\frac{d y}{d x} + 3 x^{2} y = x

x \to 1 lim (x^{4} - x^{2} + 2)

y^{^{'}} + 0.4 \cdot y = 0.7 \cdot (299 \cdot e^{- 0.7 \cdot x})

What is the integral of 1/(sqrt(16-49x^2)) ?
The integral of 1/(sqrt(16-49x^2)) is 1/7 arcsin(7/4 x)+C