Can you do integral of sec^5(x)tan^2(x) without reduction formula?

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SUMMARY

The integral of sec5(x)tan2(x) can be computed without a reduction formula by transforming the integrand into terms of sec(x). The expression can be rewritten as ∫sec7(x)dx - ∫sec5(x)dx. Integration by parts is applied to both integrals, with u = sec5(x) and dv = sec2(x)dx for the first integral, and u = sec3(x) and dv = sec2(x)dx for the second integral. This method effectively simplifies the problem and allows for a solution.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with trigonometric identities, particularly secant and tangent functions.
  • Knowledge of integral calculus, including the manipulation of integrals.
  • Experience with algebraic manipulation of trigonometric expressions.
NEXT STEPS
  • Study the method of integration by parts in detail.
  • Explore trigonometric identities and their applications in integration.
  • Practice solving integrals involving secant and tangent functions.
  • Learn about reduction formulas for integrals of trigonometric functions.
USEFUL FOR

Students studying calculus, particularly those focusing on integral calculus, as well as educators seeking to enhance their teaching methods for trigonometric integrals.

aselin0331
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Homework Statement



Can you do integral of sec^5(x)tan^2(x) without reduction formula?


Homework Equations





The Attempt at a Solution



If so, would it be integration by parts? I tried splitting it up in way too many ways to post them on here.

Thanks for any hints!
 
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aselin0331 said:

Homework Statement



Can you do integral of sec^5(x)tan^2(x) without reduction formula?


Homework Equations





The Attempt at a Solution



If so, would it be integration by parts? I tried splitting it up in way too many ways to post them on here.

Thanks for any hints!
First, get everything in terms of sec(x).
\int sec^5(x)tan^2(x)dx = \int sec^7(x) - sec^5(x)dx = \int sec^7(x)dx - \int sec^5(x)dx

Use integration by parts on both integrals on the right. For the first u = sec5(x) and dv = sec2(x) dx. For the second integral, u = sec3(x) and dv = sec2(x) dx.

I haven't taken this all the way through, but I'm reasonably sure it will work. You will probably need to solve for your integral algebraically.
 
You're fantastic! I was staring at Sec^7 and sec^5 for a while and that just clicked in!

Thanks
 

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