What is the integral of sqrt(1-7x^2) ?

The integral of sqrt(1-7x^2) is 1/(2sqrt(7))(arcsin(sqrt(7)x)+1/2 sin(2arcsin(sqrt(7)x)))+C

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integral of sqrt(1-7x^2)

\int 1 - 7 x^{2} d x

\frac{1}{2 7} (arcsin (7 x) + \frac{1}{2} sin (2 arcsin (7 x))) + C

Solution steps

\int 1 - 7 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{7} cos^{2} (u) d u

= \int \frac{1}{7} cos^{2} (u) d u

Take the constant out:

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

= \frac{1}{7} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= \frac{1}{7} \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

= \frac{1}{7} \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

= \frac{1}{7} \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)

= \frac{1}{7} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (7 x)

= \frac{1}{7} \cdot \frac{1}{2} (arcsin (7 x) + \frac{1}{2} sin (2 arcsin (7 x)))

Simplify

\frac{1}{7} \cdot \frac{1}{2} (arcsin (7 x) + \frac{1}{2} sin (2 arcsin (7 x))) : \frac{1}{2 7} (arcsin (7 x) + \frac{1}{2} sin (2 arcsin (7 x)))

= \frac{1}{2 7} (arcsin (7 x) + \frac{1}{2} sin (2 arcsin (7 x)))

Add a constant to the solution

= \frac{1}{2 7} (arcsin (7 x) + \frac{1}{2} sin (2 arcsin (7 x))) + C

n = 2 \sum \infty \frac{5}{2 ^{n}}

t e^{- 5 t}

5 t ln (t)

\int_{0}^{b} \frac{1}{16 - x ^{2}} d x

What is the integral of sqrt(1-7x^2) ?
The integral of sqrt(1-7x^2) is 1/(2sqrt(7))(arcsin(sqrt(7)x)+1/2 sin(2arcsin(sqrt(7)x)))+C