What is the integral of 5/(sqrt(36-25x^2)) ?

The integral of 5/(sqrt(36-25x^2)) is arcsin(5/6 x)+C

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integral of 5/(sqrt(36-25x^2))

\int \frac{5}{36 - 25 x ^{2}} d x

arcsin (\frac{5}{6} x) + C

Solution steps

\int \frac{5}{36 - 25 x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{36 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= 5 \cdot \int \frac{1}{5} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{5} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{5}{6} x)

= 5 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{6} x)

Simplify

5 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{6} x) : arcsin (\frac{5}{6} x)

= arcsin (\frac{5}{6} x)

Add a constant to the solution

= arcsin (\frac{5}{6} x) + C

x \to + 0 lim (\frac{sin ( + ( x ) )}{x})

\int sin^{4} (x) cos^{6} (x) d x

y^{^{'}} = - \frac{4 t}{y}, y (0) = 2

\frac{\partial}{\partial x} (\frac{( x + 1 )}{x ^{2}})

\frac{\partial}{\partial x} (\frac{2}{x ^{2} + y ^{2} + z ^{2}})

What is the integral of 5/(sqrt(36-25x^2)) ?
The integral of 5/(sqrt(36-25x^2)) is arcsin(5/6 x)+C