What is the integral of 7/(sqrt(25-x^2)) ?

The integral of 7/(sqrt(25-x^2)) is 7arcsin(1/5 x)+C

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integral of 7/(sqrt(25-x^2))

\int \frac{7}{25 - x ^{2}} d x

7 arcsin (\frac{1}{5} x) + C

Solution steps

\int \frac{7}{25 - x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 7 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 7 \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{1}{5} x)

= 7 \cdot 1 \cdot arcsin (\frac{1}{5} x)

Simplify

= 7 arcsin (\frac{1}{5} x)

Add a constant to the solution

= 7 arcsin (\frac{1}{5} x) + C

2 x y

y^{^{'}} = y + sin (x) - cos (x)

f (x) = \frac{7 x - 5}{x ^{2} + 4 x}

f (x) = x^{4} + 4 x^{3} - 48 x^{2} + 6

\int_{0}^{9} 2 x d x

What is the integral of 7/(sqrt(25-x^2)) ?
The integral of 7/(sqrt(25-x^2)) is 7arcsin(1/5 x)+C