# How do you find the Integral of #(1-x^2)^(3/2)/ (x^6)#?

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To find the integral of (1-x^2)^(3/2)/x^6, we can use trigonometric substitution. Let's make the substitution x = sin(θ):

- Find dx in terms of dθ: dx = cos(θ) dθ.
- Substitute x = sin(θ) and dx = cos(θ) dθ into the integral.

The integral becomes:

∫(1-sin^2(θ))^(3/2)/sin^6(θ) * cos(θ) dθ.

Using the trigonometric identity 1 - sin^2(θ) = cos^2(θ), we simplify the integrand to:

∫cos^3(θ)/sin^6(θ) * cos(θ) dθ = ∫cos^4(θ)/sin^6(θ) dθ.

Using the double-angle identity for cosine, cos^2(θ) = (1 + cos(2θ))/2, we can express cos^4(θ) in terms of cos(2θ):

cos^4(θ) = (1 + cos(2θ))^2/4 = (1 + 2cos(2θ) + cos^2(2θ))/4.

Now, we substitute this expression back into the integral and simplify:

∫(1 + 2cos(2θ) + cos^2(2θ))/4sin^6(θ) dθ.

Using the half-angle formula cos^2(θ) = (1 + cos(2θ))/2, we simplify cos^2(2θ) to (1 + cos(4θ))/2:

∫(1 + 2cos(2θ) + (1 + cos(4θ))/2)/4sin^6(θ) dθ.

Simplifying further:

∫(3/4 + 1/2cos(2θ) + 1/8cos(4θ))/sin^6(θ) dθ.

Now, we can integrate each term separately. The integral of 3/4sin^6(θ) is (3/4)∫csc^6(θ) dθ, which can be evaluated using a reduction formula. The other two terms involve integrating cosines, which can be done using standard trigonometric integral formulas.

After integrating each term, substitute back sin(θ) = x to get the final result in terms of x.

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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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