What is the integral of 1/(sqrt(-u^2+b-a)) ?

The integral of 1/(sqrt(-u^2+b-a)) is arcsin(u/(sqrt(b-a)))+C

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integral of 1/(sqrt(-u^2+b-a))

\int \frac{1}{- u ^{2} + b - a} d u

arcsin (\frac{u}{b - a}) + C

Solution steps

\int \frac{1}{- u ^{2} + b - a} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

v = arcsin (\frac{1}{b - a} u)

= 1 \cdot arcsin (\frac{1}{b - a} u)

Simplify

1 \cdot arcsin (\frac{1}{b - a} u) : arcsin (\frac{u}{b - a})

= arcsin (\frac{u}{b - a})

Add a constant to the solution

= arcsin (\frac{u}{b - a}) + C

x \to \infty lim (\frac{x - x x}{2 x ^{\frac{3}{2}} + 3 x - 5})

9 x + 10

x \to 0 + lim ((x)^{4 x})

\int (\frac{x}{x - 1}) d x

x \to θ lim (x^{2} - 5)

What is the integral of 1/(sqrt(-u^2+b-a)) ?
The integral of 1/(sqrt(-u^2+b-a)) is arcsin(u/(sqrt(b-a)))+C