find-x-2-2-x-2-x-dx-

Question Number 98182 by abdomathmax last updated on 12/Jun/20

find \int x^{2} \sqrt{\frac{2 - x}{2 + x}} dx

Answered by MJS last updated on 12/Jun/20

\int x^{2} \sqrt{\frac{2 - x}{2 + x}} d x =

[t = \sqrt{\frac{2 + x}{2 - x}} \to d x = \frac{1}{2} \sqrt{(2 + x) {(2 - x)}^{3}} d t]

= 32 \int \frac{{(t^{2} - 1)}^{2}}{{(t^{2} + 1)}^{4}} d t =

[Ostrogradski]

= \frac{8 t (3 t^{4} + 4 t^{2} + 9)}{3 {(t^{2} + 1)}^{3}} + 8 \int \frac{d t}{t^{2} + 1} =

= \frac{8 t (3 t^{4} + 4 t^{2} + 9)}{3 {(t^{2} + 1)}^{3}} + 8 \arctan t =

= \frac{1}{3} (x^{2} - 3 x + 8) \sqrt{4 - x^{2}} + 8 \arctan \frac{\sqrt{2 + x}}{\sqrt{2 - x}} =

= \frac{1}{3} (x^{2} - 3 x + 8) \sqrt{4 - x^{2}} + 8 \arcsin \frac{\sqrt{2 + x}}{2} + C

Commented by mathmax by abdo last updated on 12/Jun/20

thank you sir .

Answered by 1549442205 last updated on 12/Jun/20

Putting x = 2 \cos φ (φ \in [0; π]) \Rightarrow dx = - 2 \sin φ d φ

F = \int {4 \cos}^{2} φ . \tan \frac{φ}{2} (- 2 \sin φ) d φ . Put \tan \frac{φ}{2} = u

\Rightarrow du = \frac{1}{2} (1 + u^{2}) d φ, \sin φ = \frac{2 u}{1 + u^{2}}, \cos φ = \frac{1 - u^{2}}{1 + u^{2}}

F = - 8 \int {(\frac{1 - u^{2}}{1 + u^{2}})}^{2} . u . \frac{2 u}{1 + u^{2}} . \frac{2 du}{1 + u^{2}} = - 32 \int \frac{u^{2} {(1 - u^{2})}^{2}}{{(1 + u^{2})}^{4}} du

= - 32 \int \frac{u^{6} - {2 u}^{4} + u^{2}}{{(1 + u^{2})}^{4}} du = - 32 \int \frac{{(u^{2} + 1)}^{3} - 5 {(u^{2} + 1)}^{2} + 8 (u^{2} + 1) - 4}{{(u^{2} + 1)}^{4}} du

= - 32 \int \frac{du}{u^{2} + 1} + 160 \int \frac{du}{{(u^{2} + 1)}^{2}} - 96 \int \frac{du}{{(u^{2} + 1)}^{3}} + 128 \int \frac{du}{{(u^{2} + 1)}^{4}} = A + B + C + D

Note that we have the following formula :

I_{n} = \int \frac{dt}{{(t^{2} + a^{2})}^{n}} = \frac{1}{{2 a}^{2} (n - 1)} . \frac{t}{{(t^{2} + a^{2})}^{n - 1}} + \frac{1}{a^{2}} . \frac{2 n - 3}{2 n - 2} . I_{n - 1}

Hence, B = {160 I}_{2} = 160 {\frac{1}{2} . \frac{u}{u^{2} + 1} + \frac{1}{2} \int \frac{du}{u^{2} + 1}} = \frac{80 u}{u^{2} + 1} + 80 \arctan (u)

C = - {96 I}_{3} = - 96 {\frac{1}{4} . \frac{u}{{(u^{2} + 1)}^{2}} + \frac{3}{4} [\frac{u}{2 (u^{2} + 1)} + \frac{1}{2} \arctan (u)]}

= \frac{- 24 u}{{(u^{2} + 1)}^{2}} - \frac{36 u}{u^{2} + 1} - 36 \arctan (u)

D = {128 I}_{4} = 128 {\frac{1}{6} . \frac{u}{{(u^{2} + 1)}^{3}} + \frac{5}{6} I_{3}} = 128 {\frac{u}{6 {(u^{2} + 1)}^{3}} + \frac{5}{6} [\frac{u}{4 {(u^{2} + 1)}^{2}} + (\frac{3 u}{8 (u^{2} + 1)} + \frac{3}{8} \arctan (u))]}

= \frac{64 u}{3 {(u^{2} + 1)}^{3}} + \frac{80 u}{3 {(u^{2} + 1)}^{2}} + \frac{80 u}{3 (u^{2} + 1)} + \frac{120}{3} \arctan (u)

Thus, F = - 32 \arctan (u) + \frac{80 u}{u^{2} + 1} + 80 \arctan (u) - \frac{24 u}{{(u^{2} + 1)}^{2}}

- \frac{36 u}{u^{2} + 1} - 36 \arctan (u) + \frac{64 u}{3 {(u^{2} + 1)}^{3}} + \frac{80 u}{3 {(u^{2} + 1)}^{2}} + \frac{80 u}{3 (u^{2} + 1)} + \frac{120}{3} \arctan (u)

= \frac{156}{3} \arctan (u) + \frac{224 u}{3 (u^{2} + 1)} + \frac{152 u}{3 {(u^{2} + 1)}^{2}} + \frac{64 u}{3 {(u^{2} + 1)}^{3}} with u = \tan \frac{φ}{2} and x = 2 \cos φ

Commented by mathmax by abdo last updated on 12/Jun/20

thankx sir .