What is the integral from 0 to a of sqrt(2x-x^2) ?

The integral from 0 to a of sqrt(2x-x^2) is 1/2 (arcsin(a-1)+pi/2+1/2 sin(2arcsin(a-1)))

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integral from 0 to a of sqrt(2x-x^2)

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

\frac{1}{2} (arcsin (a - 1) + \frac{π}{2} + \frac{1}{2} sin (2 arcsin (a - 1)))

Solution steps

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

Complete the square

2 x - x^{2} : - (x - 1)^{2} + 1

= \int_{0}^{a} - (x - 1)^{2} + 1 d x

Apply u-substitution

: \int_{- 1}^{a - 1} - u^{2} + 1 d u

= \int_{- 1}^{a - 1} - u^{2} + 1 d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{a r c s i n (a - 1)} cos^{2} (v) d v

= \int_{- \frac{π}{2}}^{a r c s i n (a - 1)} cos^{2} (v) d v

Rewrite using trig identities

= \int_{- \frac{π}{2}}^{a r c s i n (a - 1)} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{a r c s i n (a - 1)} 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{- \frac{π}{2}}^{a r c s i n (a - 1)} 1 d v + \int_{- \frac{π}{2}}^{a r c s i n (a - 1)} cos (2 v) d v)

\int_{- \frac{π}{2}}^{a r c s i n (a - 1)} 1 d v = arcsin (a - 1) + \frac{π}{2}

\int_{- \frac{π}{2}}^{a r c s i n (a - 1)} cos (2 v) d v = \frac{1}{2} sin (2 arcsin (a - 1))

= \frac{1}{2} (arcsin (a - 1) + \frac{π}{2} + \frac{1}{2} sin (2 arcsin (a - 1)))

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

\int_{- 2 x}^{x} \frac{1}{csc ^{5} ( t )} d t

\int_{0}^{2} 5 x^{2} - 8 x + 4 d x

\int_{1}^{5} \frac{2}{t ^{3}} d t

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

What is the integral from 0 to a of sqrt(2x-x^2) ?
The integral from 0 to a of sqrt(2x-x^2) is 1/2 (arcsin(a-1)+pi/2+1/2 sin(2arcsin(a-1)))