What is the integral from 3 to 6 of 1/(sqrt(6x-x^2)) ?

The integral from 3 to 6 of 1/(sqrt(6x-x^2)) is pi/2

Popular >

integral from 3 to 6 of 1/(sqrt(6x-x^2))

\int_{3}^{6} \frac{1}{6 x - x ^{2}} d x

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{3}^{6} \frac{1}{6 x - x ^{2}} d x

Complete the square

6 x - x^{2} : - (x - 3)^{2} + 9

= \int_{3}^{6} \frac{1}{- ( x - 3 ) ^{2} + 9} d x

Apply u-substitution

: \int_{0}^{3} \frac{1}{- u ^{2} + 9} d u

= \int_{0}^{3} \frac{1}{- u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 1 d v

= \int_{0}^{\frac{π}{2}} 1 d v

Integral of a constant:

\int a d x = a x

= [1 \cdot v]_{0}^{\frac{π}{2}}

Simplify

= [v]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\int_{1}^{2} x^{2} x^{3} + 1 d x

\int_{1}^{2} \frac{1}{1 - e ^{x}} d x

\int_{0}^{L} sin (\frac{nπ x}{L}) d x

\int_{0}^{1} \frac{1}{4 + x} d x

\int_{1}^{4} (x^{2} - 9 x) d x

What is the integral from 3 to 6 of 1/(sqrt(6x-x^2)) ?
The integral from 3 to 6 of 1/(sqrt(6x-x^2)) is pi/2