How do you integrate #int x^3*(1-x^4)^6 dx #?

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Answer 1 · 2018-04-05T16:47:15

#intx^3(1-x^4)^6dx=-1/28(1-x^4)^7+c#

#intx^3(1-x^4)^6dx---(1)#

method 1: inspection

we note that the function outside the bracket is a multiple of the bracket differentiated, so we can approach this integral by inspection.

We will 'guess' the bracket to the power +1, differentiate and compare it with the required integral.

#d/(dx)(1-x^4)^7=7(1-x^4)^6xx(-4x^3)#

#=-28x^3(1-x^4)^6--(2)#

comparing #(1) " & "(2)# we can conclude

#intx^3(1-x^4)^6dx=-1/28(1-x^4)^7+c#

method 2 : substitution

#intx^3(1-x^4)^6dx---(1)#

#u=1-x^4=>du=-4x^3dx#

#(1)rarrintcancel(x^3)u^6xx-1/(4cancel(x^3))du#

#=-1/4intu^6du=-1/4xxu^7/7+c#

#=-1/28u^7+c=-1/28(1-x^4)^7+c#