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

Answer 1

Charles Anctil

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#

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Answer 2

Scarlett Allison

To integrate (\int \frac{1}{(4-x)^3} , dx) using trigonometric substitution, you can make the substitution (x = 4 - \tan^2(\theta)), which will lead to a form that can be integrated using trigonometric identities. After making this substitution, you can then express (dx) in terms of (\theta), and rewrite the integral in terms of (\theta). Finally, you can integrate with respect to (\theta), then substitute back the original variable (x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.