What is the integral of xsqrt(2-\sqrt{1-x^2)} ?

The integral of xsqrt(2-\sqrt{1-x^2)} is -2/5 (2-sqrt(1-x^2))^{5/2}+4/3 (2-sqrt(1-x^2))^{3/2}+C

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integral of xsqrt(2-\sqrt{1-x^2)}

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

- \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int - u 2 - u d u

= \int - u 2 - u d u

Take the constant out:

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

= - \int u 2 - u d u

Apply u-substitution

: \int - (- v + 2) v d v

= - \int - (- v + 2) v d v

Take the constant out:

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

= - (- \int (- v + 2) v d v)

Expand

(- v + 2) v : - v^{\frac{3}{2}} + 2 v

= - (- \int - v^{\frac{3}{2}} + 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

= - (- (- \int v^{\frac{3}{2}} d v + \int 2 v d v))

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int 2 v d v = \frac{4}{3} v^{\frac{3}{2}}

= - (- (- \frac{2}{5} v^{\frac{5}{2}} + \frac{4}{3} v^{\frac{3}{2}}))

Substitute back

= - (- (- \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}}))

Simplify

= - \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}} + C

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

y^{^{'}} + 6 y = e^{x}

y = x^{2} 16 - x^{2}

\int_{0}^{\frac{π}{6}} 6 sec^{3} (x) d x

sin (6 (t - \frac{π}{6}))

What is the integral of xsqrt(2-\sqrt{1-x^2)} ?
The integral of xsqrt(2-\sqrt{1-x^2)} is -2/5 (2-sqrt(1-x^2))^{5/2}+4/3 (2-sqrt(1-x^2))^{3/2}+C