What is the integral from 3 to x of sqrt(9-y^2) ?

The integral from 3 to x of sqrt(9-y^2) is 9/2 (arcsin(x/3)-pi/2+1/2 sin(2arcsin(x/3)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 3 to x of sqrt(9-y^2)

\int_{3}^{x} 9 - y^{2} d y

\frac{9}{2} (arcsin (\frac{x}{3}) - \frac{π}{2} + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

Solution steps

\int_{3}^{x} 9 - y^{2} d y

Apply Trigonometric Substitution

: \int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} 9 cos^{2} (u) d u

= \int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} 9 cos^{2} (u) d u

Take the constant out:

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

= 9 \cdot \int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} 1 + cos (2 u) d u

Apply the Sum Rule:

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

= 9 \cdot \frac{1}{2} (\int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} 1 d u + \int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} cos (2 u) d u)

\int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} 1 d u = arcsin (\frac{x}{3}) - \frac{π}{2}

\int_{\frac{π}{2}}^{a r c s i n (\frac{x}{3})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{x}{3}))

= 9 \cdot \frac{1}{2} (arcsin (\frac{x}{3}) - \frac{π}{2} + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{x}{3}) - \frac{π}{2} + \frac{1}{2} sin (2 arcsin (\frac{x}{3}))) : \frac{9}{2} (arcsin (\frac{x}{3}) - \frac{π}{2} + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

= \frac{9}{2} (arcsin (\frac{x}{3}) - \frac{π}{2} + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

\int_{0}^{6} 3 d x

\int_{\frac{1}{2}}^{4} x^{3} - 6 x^{2} + 9 x + 1 d x

\int_{- 2}^{2} e^{u + 2} d u

\int_{0}^{x} 16 - t^{2} d t

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

What is the integral from 3 to x of sqrt(9-y^2) ?
The integral from 3 to x of sqrt(9-y^2) is 9/2 (arcsin(x/3)-pi/2+1/2 sin(2arcsin(x/3)))