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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} = - y sin (2 x)

y = 4 e^{x} cos (x), 4, at x = 0

\frac{d y}{d x} y = 4 x^{2} + 2

\frac{5}{s}

\int_{1}^{6} \frac{x}{2} + \frac{1}{2 x} d x

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