# Trig substitution into integrals

In summary, the conversation discusses testing for convergence of a series and using the integral test with a trig substitution. The question arises about whether the substitution is valid when the upper bound of infinity fits inside the sine function. It is determined that the substitution is possible, but it may be simpler to use partial fractions instead. The conversation also mentions considering a more appropriate trig substitution, such as secant, and using partial fractions as the preferred technique for this integral.
I was testing for convergence of a series:
∑$\frac{1}{n^2 -1}$ from n=3 to infinity

I used the integral test, substituting n as 2sin(u)

so here's the question:
when using the trig substitution, I realized the upperbound, infinity, would fit inside the sine.

Is it still possible to make the substitution? Or is there a restriction when this happens?

I was testing for convergence of a series:
∑$\frac{1}{n^2 -1}$ from n=3 to infinity

I used the integral test, substituting n as 2sin(u)

so here's the question:
when using the trig substitution, I realized the upperbound, infinity, would fit inside the sine.
What does "fit inside the sine" mean?
Is it still possible to make the substitution? Or is there a restriction when this happens?
Sure, you can make the substitution. The integral will be from 3 to, say b, and you take the limit as b → ∞.

Not that you asked, but it's probably simpler and quicker to break up 1/(n2 - 1) using partial fractions.

Inside the sine meaning, the argument of the 'arcsine' would only range from -1 to 1.
So I'm guessing you can't make the substitution because arcsin(infinity) = error?

If you're looking for an appropriate trig substitution for the definite integral (and not just one that gets you a correct antidierivative), then ##\sec u## is the way to go. But like Mark44 said, partial fractions is really the "right" technique of integration for this particular integral.

Yes, it is still possible to make the substitution. The substitution of n as 2sin(u) is a valid substitution for this integral, regardless of the value of the upper bound. The upper bound being infinity does not affect the validity of the substitution. However, it is important to keep in mind that when evaluating the integral, the upper bound must also be substituted as infinity. This may require some additional algebraic manipulation to simplify the integral before evaluating it at infinity.

## 1. What is "trig substitution" in integrals?

Trig substitution is a technique used to simplify integrals involving expressions with square roots or quadratic terms. It involves substituting a trigonometric function for a variable in the integral, which allows for the use of trigonometric identities to simplify the integral.

## 2. When should I use trig substitution in integrals?

Trig substitution should be used when the integral contains expressions with square roots or quadratic terms, and cannot be solved using basic integration techniques. It is also useful when the integral contains a combination of trigonometric functions.

## 3. How do I choose the appropriate trig substitution?

The choice of trig substitution depends on the form of the integral. For integrals involving expressions with the form x2 + a2, use a substitution of x = a tanθ. For integrals involving expressions with the form x2 - a2, use a substitution of x = a secθ. For integrals involving expressions with the form a2 - x2, use a substitution of x = a sinθ or x = a cosθ.

## 4. What are the common trig identities used in trig substitution?

The most commonly used trig identities in trig substitution are: sin2θ + cos2θ = 1, tan2θ + 1 = sec2θ, 1 + cot2θ = csc2θ. These identities are used to simplify the integral and express it in terms of a single trigonometric function.

## 5. Are there any special cases to consider when using trig substitution?

Yes, there are two special cases to consider when using trig substitution: when the integral contains an odd power of secθ, and when the integral contains an even power of secθ. In these cases, a different substitution or manipulation may be needed in order to simplify the integral.

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