What is the integral of 1/(sqrt(20+8x-x^2)) ?

The integral of 1/(sqrt(20+8x-x^2)) is arcsin(1/6 (x-4))+C

Popular >

integral of 1/(sqrt(20+8x-x^2))

\int \frac{1}{20 + 8 x - x ^{2}} d x

arcsin (\frac{1}{6} (x - 4)) + C

Solution steps

\int \frac{1}{20 + 8 x - x ^{2}} d x

Complete the square

20 + 8 x - x^{2} : - (x - 4)^{2} + 36

= \int \frac{1}{- ( x - 4 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 36} d u

= \int \frac{1}{- u ^{2} + 36} 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

= 1 \cdot arcsin (\frac{1}{6} (x - 4))

Simplify

= arcsin (\frac{1}{6} (x - 4))

Add a constant to the solution

= arcsin (\frac{1}{6} (x - 4)) + C

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

\frac{s + 5}{s ^{2} + 6 s + 18}

y^{^{''}} - 6 y^{^{'}} + 9 y = 6 x^{2} + 2 - 12 e^{3 x}

(3 + sin (2 x))^{2}, at x = 0

\frac{2}{x ^{2} - x}

What is the integral of 1/(sqrt(20+8x-x^2)) ?
The integral of 1/(sqrt(20+8x-x^2)) is arcsin(1/6 (x-4))+C