What is the integral of 1/(sqrt(196-x^2)) ?

The integral of 1/(sqrt(196-x^2)) is arcsin(1/14 x)+C

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integral of 1/(sqrt(196-x^2))

\int \frac{1}{196 - x ^{2}} d x

arcsin (\frac{1}{14} x) + C

Solution steps

\int \frac{1}{196 - x ^{2}} d x

Apply Trigonometric Substitution

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = arcsin (\frac{1}{14} x)

= 1 \cdot arcsin (\frac{1}{14} x)

Simplify

= arcsin (\frac{1}{14} x)

Add a constant to the solution

= arcsin (\frac{1}{14} x) + C

\int (5 - 2 x)^{9} d x

x \to \infty lim (x sin (\frac{2 π}{x}))

\frac{d}{d x} (12 - 3 x - 4 y)

\int \frac{14}{( 49 x ^{2} + 1 ) ^{2}} d x

d er i v a t i v e f (x) = 2 x^{2} - 9 x + 10

What is the integral of 1/(sqrt(196-x^2)) ?
The integral of 1/(sqrt(196-x^2)) is arcsin(1/14 x)+C