What is the integral of sqrt(8-y^2) ?

The integral of sqrt(8-y^2) is 4(arcsin(1/(2sqrt(2))y)+1/2 sin(2arcsin(1/(2sqrt(2))y)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(8-y^2)

\int 8 - y^{2} d y

4 (arcsin (\frac{1}{2 2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 2} y))) + C

Solution steps

\int 8 - y^{2} d y

Apply Trigonometric Substitution

: \int 8 cos^{2} (u) d u

= \int 8 cos^{2} (u) d u

Take the constant out:

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

= 8 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 8 \cdot \int \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

= 8 \cdot \frac{1}{2} \cdot \int 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

= 8 \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 8 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{2 2} y)

= 8 \cdot \frac{1}{2} (arcsin (\frac{1}{2 2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 2} y)))

Simplify

8 \cdot \frac{1}{2} (arcsin (\frac{1}{2 2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 2} y))) : 4 (arcsin (\frac{1}{2 2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 2} y)))

= 4 (arcsin (\frac{1}{2 2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 2} y)))

Add a constant to the solution

= 4 (arcsin (\frac{1}{2 2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 2} y))) + C

\int \frac{e ^{x} + e ^{2 x} + e ^{3 x}}{e ^{4 x}} d x

\int - \frac{7}{x ^{2}} d x

\int \frac{x ^{2}}{3 ( 1 + x ^{2} )} d x

\int x 7 x^{2} + 5 d x

\int ln (2) x^{2} d x

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