What is the integral of sqrt(16-(x-1)^2) ?

The integral of sqrt(16-(x-1)^2) is 8(arcsin(1/4 (x-1))+1/2 sin(2arcsin(1/4 (x-1))))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(16-(x-1)^2)

\int 16 - (x - 1)^{2} d x

8 (arcsin (\frac{1}{4} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 1)))) + C

Solution steps

\int 16 - (x - 1)^{2} d x

Apply u-substitution

= \int 16 - u^{2} d u

Apply Trigonometric Substitution

= \int 16 cos^{2} (v) d v

Take the constant out:

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

= 16 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 16 \cdot \int \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

= 16 \cdot \frac{1}{2} \cdot \int 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

= 16 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= 16 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 1))))

Simplify

16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 1)))) : 8 (arcsin (\frac{1}{4} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 1))))

= 8 (arcsin (\frac{1}{4} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 1))))

Add a constant to the solution

= 8 (arcsin (\frac{1}{4} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 1)))) + C

\int (9^{x} - 12 e^{x} - 5 sin (x) + 8) d x

\int (x) e^{x} d x

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

\int (sin (x) + 2 cos (x))^{2} d x

\int x^{8} cos (x) d x

What is the integral of sqrt(16-(x-1)^2) ?
The integral of sqrt(16-(x-1)^2) is 8(arcsin(1/4 (x-1))+1/2 sin(2arcsin(1/4 (x-1))))+C