bemath-x-6-1-x-2-dx-

Question Number 107212 by bemath last updated on 09/Aug/20

⊚ b e m a t h ⊚

\int x^{6} \sqrt{1 - x^{2}} d x ?

Answered by Ar Brandon last updated on 09/Aug/20

I = \int x^{6} \sqrt{1 - x^{2}} dx, x = \sin θ

I = \int \sin^{6} θ \cos^{2} θ d θ = \int (\sin^{6} θ - \sin^{8} θ) d θ

{(z - \frac{1}{z})}^{6} = {(2 isin θ)}^{6}, z = \cos θ + isin θ

z^{6} - {6 z}^{5} (\frac{1}{z}) + {15 z}^{4} (\frac{1}{z^{2}}) - {20 z}^{3} (\frac{1}{z^{3}}) + {15 z}^{2} (\frac{1}{z^{4}}) - 6 z (\frac{1}{z^{5}}) + \frac{1}{z^{6}} = - {64 \sin}^{6} θ

z^{6} + \frac{1}{z^{6}} - 6 (z^{4} + \frac{1}{z^{4}}) + 15 (z^{2} + \frac{1}{z^{2}}) - 20 = - {64 \sin}^{6} θ

2 \cos 6 θ - 12 \cos 4 θ + 30 \cos 2 θ - 20 = - {64 \sin}^{6} θ

{(z - \frac{1}{z})}^{8} = {(2 isin θ)}^{8}

1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1

z^{8} + \frac{1}{z^{8}} - 8 (z^{6} + \frac{1}{z^{6}}) + 28 (z^{4} + \frac{1}{z^{4}}) - 56 (z^{2} + \frac{1}{z^{2}}) + 70 = {256 \sin}^{8} θ

2 \cos 8 θ - 16 \cos 6 θ + 56 \cos 4 θ - 112 \cos 2 θ + 70 = {256 \sin}^{8} θ

\Rightarrow I = - \frac{1}{64} [\frac{2}{6} \sin 6 θ - \frac{12}{4} \sin 4 θ + \frac{30}{2} \sin 2 θ - 20 θ]

- \frac{1}{256} [\frac{2}{8} \sin 8 θ - \frac{16}{6} \sin 6 θ + \frac{56}{4} \sin 4 θ - \frac{112}{2} \sin 2 θ + 70 θ] + C

I = \frac{1}{256} [10 θ - 4 \sin 2 θ - 2 \sin 4 θ + \frac{4}{3} \sin 6 θ - \frac{1}{4} \sin 8 θ] + C

Answered by bobhans last updated on 09/Aug/20

I_{1} = \int - \frac{1}{2} x^{5} (- 2 x \sqrt{1 - x^{2}}) dx = \int - \frac{1}{2} x^{5} \sqrt{1 - x^{2}} d (1 - x^{2})

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

I_{2} = \int - \frac{5}{6} x^{3} (- 2 x {(1 - x^{2})}^{3 / 2}) dx

I_{2} = \int - \frac{5}{6} x^{3} {(1 - x^{2})}^{3 / 2} d (1 - x^{2})

I_{2} = - \frac{1}{3} x^{3} {(1 - x^{2})}^{5 / 2} + \int x^{2} {(1 - x^{2})}^{5 / 2} dx

I_{3} = \int - \frac{1}{2} x (- 2 x {(1 - x^{2})}^{5 / 2}) dx

I_{3} = - \frac{1}{2} x . \frac{2}{7} {(1 - x^{2})}^{7 / 2} + \int \frac{1}{7} {(1 - x^{2})}^{7 / 2}

I_{3} = - \frac{1}{7} x {(1 - x^{2})}^{7 / 2} + \int \frac{1}{7} {(1 - x^{2})}^{7 / 2} dx

Answered by 1549442205PVT last updated on 10/Aug/20

Set \frac{1}{x^{2}} - 1 = u^{2} \Rightarrow 2 udu = - \frac{2}{x^{3}} dx

\Rightarrow dx = - {ux}^{3} du, \sqrt{1 - x^{2}} = ux, x^{2} = \frac{1}{u^{2} + 1}

F = - \int x^{6} . (ux) ({ux}^{3}) du = - \int x^{10} u^{2} du

= - \int \frac{u^{2} du}{{(1 + u^{2})}^{5}} = - \int \frac{u^{2} + 1 - 1}{{(u^{2} + 1)}^{5}} du

= - \int \frac{du}{{(1 + u^{2})}^{4}} + \int \frac{du}{{(1 + u^{2})}^{5}} = I_{5} - I_{4}

Apply the current formula :

I_{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}

I_{4} = \frac{1}{6} . \frac{u}{{(u^{2} + 1)}^{3}} + \frac{5}{6} I_{3} =

\frac{u}{6 {(u^{2} + 1)}^{3}} + \frac{5}{6} [\frac{1}{4} . \frac{u}{{(u^{2} + 1)}^{2}} + \frac{3}{4} . I_{2}]

= \frac{u}{6 {(u^{2} + 1)}^{3}} + \frac{5 u}{24 {(u^{2} + 1)}^{2}} + \frac{5}{8} I_{2}

= \frac{u}{6 {(u^{2} + 1)}^{3}} + \frac{5 u}{24 {(u^{2} + 1)}^{2}} + \frac{5}{8} [\frac{1}{2} . \frac{u}{(u^{2} + 1)} + \frac{1}{2} \int \frac{du}{u^{2} + 1}]

= \frac{u}{6 {(u^{2} + 1)}^{3}} + \frac{5 u}{24 {(u^{2} + 1)}^{2}} + \frac{5 u}{16 (u^{2} + 1)} + \frac{5}{16} \tan^{- 1} (u)

I_{5} = \frac{1}{8} . \frac{u}{{(u^{2} + 1)}^{4}} + \frac{7}{8} . I_{4} . Therefore,

F = I_{5} - I_{4} = \frac{1}{8} . \frac{u}{{(u^{2} + 1)}^{4}} - \frac{1}{8} I_{4}

= \frac{u}{8 {(u^{2} + 1)}^{4}} - \frac{1}{8} [\frac{u}{6 {(u^{2} + 1)}^{3}} + \frac{5 u}{24 {(u^{2} + 1)}^{2}} + \frac{5 u}{16 (u^{2} + 1)} + \frac{5}{16} \tan^{- 1} (u)] + C

= \frac{u}{8 {(u^{2} + 1)}^{4}} - \frac{u}{48 {(u^{2} + 1)}^{3}} - \frac{5 u}{192 {(u^{2} + 1)}^{2}}

- \frac{5 u}{128 (u^{2} + 1)} - \frac{5}{128} \tan^{- 1} (u) + C

Substituting u^{2} + 1 = 1 / x^{2} and u = \frac{\sqrt{1 - x^{2}}}{x}

we get :

F = \frac{x^{7} \sqrt{1 - x^{2}}}{8} - \frac{x^{5} \sqrt{1 - x^{2}}}{48} - \frac{5 x^{3} \sqrt{1 - x^{2}}}{192}

- \frac{5 x \sqrt{1 - x^{2}}}{128} - \frac{5}{128} \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{x}) + C

Answered by mathmax by abdo last updated on 10/Aug/20

A = \int x^{6} \sqrt{1 - x^{2}} dx we do the changement x = sint \Rightarrow

A = \int \sin^{6} t \cos^{2} t dt = \int {(\sin^{2} t)}^{3} (\frac{1 + \cos (2 t)}{2}) dt

= \frac{1}{2} \int {(\frac{1 - \cos (2 t)}{2})}^{3} (1 + \cos (2 t)) dt

= \frac{1}{16} \int {(1 - \cos (2 t))}^{2} (1 - \cos^{2} (2 t)) dt

= \frac{1}{16} \int (1 - 2 \cos (2 t) + \cos^{2} (2 t)) (1 - \frac{1 + \cos (4 t)}{2}) dt

= \frac{1}{32} \int (\frac{1 + \cos (4 t)}{2} + 1 - 2 \cos (2 t)) (1 - \cos (4 t)) dt

= \frac{1}{64} \int (3 + \cos (4 t) - 4 \cos (2 t)) (1 - \cos (4 t)) dt

= \frac{1}{64} \int (3 - 3 \cos (4 t) + \cos (4 t) - \cos^{2} (4 t) - 4 \cos (2 t) + 4 \cos (2 t) \cos (4 t)) dt

= \frac{1}{64} \int (3 - 2 \cos (4 t) - \frac{1 + \cos (8 t)}{2} - 4 \cos (2 t) + 2 {\cos (6 t) + \cos (2 t)}) dt

= \frac{1}{128} \int {6 - 4 \cos (4 t) - 1 - \cos (8 t) - 8 \cos (2 t) + 4 \cos (6 t) + 4 \cos (2 t)} dt

= \frac{1}{128} \int {5 - 4 \cos (4 t) - \cos (8 t) - 4 \cos (2 t) + 4 \cos (6 t)} dt

= \frac{5}{128} t - \frac{4}{128} \int \cos (4 t) dt - \frac{1}{128} \int \cos (8 t) dt - \frac{4}{128} \int \cos (2 t) dt

+ \frac{4}{128} \int \cos (6 t) dt

= \frac{5}{128} t - \frac{1}{128} \sin (4 t) - \frac{1}{8.128} \sin (8 t) - \frac{1}{64} \sin (2 t) + \frac{4}{6.128} \sin (6 t) + C

t = arcsinx