Inverse Trig Integration - Can I use u-substitution?

In summary, the solution manual suggests adding and subtracting 6 in the numerator to express it as the complete derivative of the denominator. This allows for the use of u-substitution and makes the second part a standard integral.
  • #1
opticaltempest
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Here is the problem and the solution manual's solution to the problem. I couldn't see that I had to add 6 and subtract 6 in the numerator then split the integral into two separate integrals. Is there an easier way to do it using u-substitution? Are their alternative ways to evaluate this integral?

http://img466.imageshack.us/img466/3753/problem0lw.jpg

Thanks
 
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  • #2
From my experience, adding and subtracting 6 is the standard and easiest way to proceed.

The clue to add and subtract 6 is that, you want to express the numerator as the complete derivative of the denominator.

ie,
[tex] \frac{d}{dx} (x^2 + 6x +13) = 2x + 6 [/tex]
Which enables you to use the substitution [itex] x^2 + 6x +13 = u [/itex].

The second part, ie, [tex] -6 \int \frac{1}{x^2 + 6x +13} dx [/tex] is then a standard integral.
 

1. Can I use u-substitution for all inverse trigonometric functions?

No, u-substitution can only be used for inverse trigonometric functions that have a derivative that is a rational function. This includes inverse sine, inverse cosine, and inverse tangent functions.

2. How do I know when to use u-substitution for inverse trigonometric functions?

If you have an integral with an inverse trigonometric function in the integrand, and the derivative of that inverse trigonometric function appears elsewhere in the integrand, then u-substitution is likely an appropriate method to use.

3. Do I need to memorize the u-substitution formula for inverse trigonometric functions?

No, the u-substitution formula for inverse trigonometric functions is simply the inverse of the derivative of the corresponding trigonometric function. For example, the u-substitution formula for inverse sine is du/dx = 1/sqrt(1-x^2).

4. Can I use u-substitution for all types of inverse trigonometric integrals?

No, u-substitution is most commonly used for integrals involving inverse trigonometric functions with a rational function as the argument. It may not be applicable for integrals with more complex arguments, such as inverse trigonometric functions nested within each other.

5. Are there any special cases where u-substitution may not work for inverse trigonometric integrals?

Yes, there are some special cases where u-substitution may not work, such as when the integral involves a trigonometric function raised to a power (e.g. sin^2(x)) or when the integral involves a combination of inverse trigonometric functions (e.g. arctan(arcsin(x))). In these cases, other methods such as trigonometric identities or integration by parts may be more appropriate.

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