use substition and use an appropriate reduction formulae

Here is the calculation with substitution x = sin t.

∫x^6 (1-x^2) dx, {from 0 to 1}
= ∫(sin t)^6 [1 - (sin t)^2] cos t dt, {from 0 to pi/2}
= ∫(sin t)^6 (cos t)^3 dt, {from 0 to pi/2} ................................. (1)

Construct a reduction formula:
For positive integers n and m,
∫(sin t)^n (cos t)^m dt, {from 0 to pi/2}
= -1/(m+1)∫(sin t)^(n-1) d[(cos t)^(m+1)], {from 0 to pi/2}
= -1/(m+1) • [ (sin t)^(n-1) • (cos t)^(m+1) ], {from 0 to pi/2}
　+ 1/(m+1)∫(cos t)^(m+1) d[(sin t)^(n-1)], {from 0 to pi/2}
= 0 + (n-1)/(m+1)∫(sin t)^(n-2) (cos t)^(m+2) dt, {from 0 to pi/2}.
i.e.
I(n, m) = (n-1)/(m+1) I(n-2, m+2),
　where I(n, m) =∫(sin t)^n (cos t)^m dt, {from 0 to pi/2},
　for non-zero integers n and m.

It follows from (1) above that,
∫x^6 (1-x^2) dx, {from 0 to 1}
= ∫(sin t)^6 (cos t)^3 dt, {from 0 to pi/2} ................................. (1)
= I(6, 3)
= 5/4 I(4, 5)
= (5/4) (3/6) I(2, 7)
= (5/4) (3/6) (1/8) I(0, 9)
= (5/64)∫(cos t)^9 dt, {from 0 to pi/2}
= (5/64)∫(cos t)^8 d(sin t), {from 0 to pi/2}
= (5/64)∫[1 - (sin t)^2]^4 d(sin t), {from 0 to pi/2}
= (5/64)∫[1 - 4(sin t)^2 + 6(sin t)^4 - 4(sin t)^6 + (sin t)^8] d(sin t), {from 0 to pi/2}
= (5/64) [sin t - 4/3 (sin t)^3 + 6/5 (sin t)^5 - 4/7 (sin t)^7 + 1/9 (sin t)^9], {from 0 to pi/2}
= (5/64) [1 - 4/3 + 6/5 - 4/7 + 1/9], {from 0 to pi/2}
= (5/64) (315 - 420 + 378 - 180 + 35)/315
= (1/64) (128)/63
= 2/63.

2007-01-28 09:21:27 補充：
In fact, the calculation would be far more straight-forward and quicker if substitution is not used.∫x^6(1-x^2) dx, {0,1}=∫(x^6-x^8) dx, {0,1}= (x^7)/7-(x^9)/9, {0,1}= 1/7 - 1/9= 2/63.