# How do you integrate #int -x^3/sqrt(144+x^2)dx# using trigonometric substitution?

The answer is

There is no need for trigonometric substitution

The integral is

Therefore,

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To integrate ( \int \frac{-x^3}{\sqrt{144 + x^2}} , dx ) using trigonometric substitution, let ( x = 12 \tan(\theta) ). Then, ( dx = 12 \sec^2(\theta) , d\theta ).

Substituting into the integral gives:

[ \int \frac{-x^3}{\sqrt{144 + x^2}} , dx = \int \frac{- (12 \tan(\theta))^3}{\sqrt{144 + (12 \tan(\theta))^2}} \cdot 12 \sec^2(\theta) , d\theta ]

Simplify the expression inside the integral, and then you can proceed to integrate it.

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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.

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.

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.

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