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Note that for sec²(x) - tan²(x) = 1, we have:

-tan²(x) = 1 - sec²(x)

tan²(x) = sec²(x) - 1

Rewrite the expression as:

∫ (sec²(x) - 1) dx

= ∫ sec²(x) dx - ∫ 1 dx

Finally, integrate each expression to get:

tan(x) - x + K where K is the arbitrary constant

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โˆ™ 2013-03-05 16:37:40
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Algebra

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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: What is the integral of tan squared x?
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