Can You Verify the Integral of Secant Squared Over (1+Tangent)^3?

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Discussion Overview

The discussion revolves around the verification of the integral of secant squared over the cube of one plus tangent. Participants explore the integration process, including substitution and the resulting expressions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the integral and proposes a substitution with \( u = 1 + \tan(t) \), leading to \( du = \sec^2(t) dt \).
  • Another participant confirms the substitution and expresses agreement with the initial steps taken.
  • A participant calculates the integral of \( \frac{1}{u^3} \) and provides a resulting expression, questioning whether a different answer was obtained from a TI calculator.
  • Another participant notes a potential error regarding a lost minus sign in the final answer but confirms that the derivative of the provided expression returns to the original integral.

Areas of Agreement / Disagreement

Participants generally agree on the substitution method and the integration steps, but there is a disagreement regarding the final expression due to a missing minus sign.

Contextual Notes

The discussion does not resolve the discrepancy regarding the final answer, and the implications of the missing minus sign remain unclarified.

karush
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$$\int\frac{\sec^2 \left({t}\right)}{\left(1+\tan\left({t}\right)\right)^3}dt$$

$$u=1+\tan\left({t}\right)\ du=\sec^2\left({t}\right) dt$$

So far ok?
 
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karush said:
$$\int\frac{\sec^2 \left({t}\right)}{\left(1+\tan\left({t}\right)\right)^3}dt$$

$$u=1+\tan\left({t}\right)\ du=\sec^2\left({t}\right) dt$$

So far ok?

Yep. (Nod)
 
$$\int\frac{1}{{u}^{3}}du =-\frac{1}{{2u}^{2}}=\frac{1}{2 (1+\tan\left({t}\right)) ^2 }+C$$

I hope the TI gave a different answer?
 
karush said:
$$\int\frac{1}{{u}^{3}}du =-\frac{1}{{2u}^{2}}=\frac{1}{2 (1+\tan\left({t}\right)) ^2 }+C$$

I hope the TI gave a different answer?

You've lost a minus sign in the final answer, but otherwise we can verify that if we take the derivative we get indeed the original integral.
 

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