What is the integral of 1/(sqrt(4-49x^2)) ?

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

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integral of 1/(sqrt(4-49x^2))

\int \frac{1}{4 - 49 x ^{2}} d x

\frac{1}{7} arcsin (\frac{7}{2} x) + C

Solution steps

\int \frac{1}{4 - 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}{2} x)

= \frac{1}{7} \cdot 1 \cdot arcsin (\frac{7}{2} x)

Simplify

= \frac{1}{7} arcsin (\frac{7}{2} x)

Add a constant to the solution

= \frac{1}{7} arcsin (\frac{7}{2} x) + C

y = e^{- k x^{2}}

1 + x^{2}

\frac{d}{d x} (e^{x^{5} + 3 x})

x^{3} - 8 x^{2}

\frac{d}{d u} (\frac{( 2 u ^{3} + 7 ) ( 3 u ^{2} - 5 )}{u ^{2} + 1})

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