# Indefinite integral approximation technique

## Homework Statement

$\int_1^\infty \! \frac{(sin(x)+5)}{x^3} \, dx$
"Find two simpler integrals, one larger and one smaller."

## The Attempt at a Solution

How could I make this a simpler (ie, solvable) integral? It's been straight forward with other integrals like this:

$\int_1^\infty \! \frac{1}{\sqrt{x^4+10}} \, dx \rightarrow \int_1^\infty \! \frac{1}{\sqrt{x^4}} \, dx$

Doing this would give an overestimate since the denominator is smaller. But how would I do it for the integral in question?

## Answers and Replies

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Can you find upper and lower bounds for sin(x)?

Can you find upper and lower bounds for sin(x)?
sin(infinity) and sin(1)? I don't understand what you mean.

Char. Limit
Gold Member
Yes, you can. Think of what y=sin(x) is restricted to...

Yes, you can. Think of what y=sin(x) is restricted to...
-1 and 1.

Char. Limit
Gold Member
Exactly. So sin(x) is always less than or equal to 1.

You can use that.

Mark44
Mentor
And so ? <= sin(x) + 5 <= ?

Filling in the question marks should make two definitie integrals (not indefinite integrals) that are a lot easier to evaluate.