How do you integrate #int 1/(4-x)^(3)dx# using trigonometric substitution?

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Answer 1 · 2017-05-02T10:03:49

The answer is #=1/(2(4-x)^2)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C#

You can perform this integral without using trigonometric substitution, we apply simple substitution

Let #u=4-x#

#du=-dx#

Therefore,

#intdx/(4-x)^3=-int(du)/u^3#

#=1/(2u^2)#

#=1/(2(4-x)^2)+C#