Question

求不定积分：∫ (cosx)^6 dx

Answer 1

求不定积分：∫cos⁶x dx

解：递推公式：∫cosⁿxdx=(1/n)(cosⁿ⁻¹xsinx)+[(n-1)/n]∫cosⁿ⁻²xdx
原式=(1/6)(cos⁵xsinx)+(5/6)∫cos⁴xdx=(1/6)(cos⁵xsinx)+(5/6)[(1/4)(cos³xsinx)+(3/4)∫cos²xdx]
=(1/6)(cos⁵xsinx)+(5/24)(cos³xsinx)+(15/24)[(1/2)(cosxsinx)+(1/2)x]+C
=(1/6)(cos⁵xsinx)+(5/24)(cos³xsinx)+(15/48)[(cosxsinx)+x]+C

Answer 2

∫ (cosx)^6 dx
=∫ (cosx)^5 dsinx
= sinx(cosx)^5 + ∫ 5(cosx)^4 (sinx)^2 dx
= sinx(cosx)^5 + ∫ 5(cosx)^4 (1-(cosx)^2) dx
6∫ (cosx)^6 dx = sinx(cosx)^5 + 5∫ (cosx)^4dx
=sinx(cosx)^5 + (5/4) (4∫ (cosx)^4dx)
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +3∫ (cosx)^2dx )
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)(2∫ (cosx)^2dx ) )
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)[sinxcosx+∫ dx] )
=sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)[sinxcosx+x] )
∫ (cosx)^6 dx = (1/6){ sinx(cosx)^5 + (5/4) ( sinx(cosx)^3 +(3/2)[sinxcosx+x] ) } + C

Answer 3