What is the integral of 7/(sqrt(4-x^2)) ?

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

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integral of 7/(sqrt(4-x^2))

\int \frac{7}{4 - x ^{2}} d x

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

Solution steps

\int \frac{7}{4 - 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}{4 - 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}{2} x)

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

Simplify

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

Add a constant to the solution

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

\frac{1}{( s - 1 ) ( s - 2 ) ( s - 3 )}

x \to 3 - lim (x 9 - x^{2})

\int tan (x) - sin (x) sin (y) d x

\int_{0}^{\frac{1}{2}} 7 arccos (x) d x

y^{^{'}} = 14 x y - 2 x

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