Primitive Function: Finding the Right Direction

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SUMMARY

The discussion focuses on solving the integral \(\int \frac{dx}{(1+\tan x)(1+\tan^2 x)}\). The user initially attempted substitutions \(t = \tan x\) and \(t = 1 + \tan x\) without success. A solution was provided using the substitution \(t = \tan x\), which simplifies the integral to \(\int \frac{dt}{(1+t)(t^{2}+1)^{2}}\), allowing for resolution through partial fractions decomposition. This method effectively addresses the integral and corrects the user's earlier missteps.

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  • Understanding of integral calculus
  • Familiarity with trigonometric functions and identities
  • Knowledge of substitution methods in integration
  • Experience with partial fractions decomposition
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  • Study advanced techniques in integral calculus
  • Learn about trigonometric substitutions in integrals
  • Explore partial fractions decomposition in detail
  • Practice solving integrals involving trigonometric functions
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Students and educators in mathematics, particularly those focusing on calculus, as well as anyone looking to enhance their skills in solving complex integrals involving trigonometric functions.

twoflower
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Hi,
I don't know how to find this primitive function:

[tex] \int \frac{dx}{(1+\tan x)(1+\tan^2 x)}[/tex]

I tried substitutions [itex]t = \tan x[/itex] or [itex]t = 1 + \tan x[/itex], but it didn't seem to help me lot...

Could someone please point me to the right direction?

Thank you.
 
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1. Set [tex]t=tan(x)\to\frac{dt}{dx}=\frac{1}{\cos^{2}(x)}=tan^{2}x+1\to{dx}=\frac{dt}{t^{2}+1}[/tex]
Thus, you've got:
[tex]\int\frac{dx}{(1+tan(x))(1+tan^{2}x)}=\int\frac{dt}{(1+t)(t^{2}+1)^{2}}[/tex]
This can be solved by partial fractions decomposition.
 
Thank you arildno, I made a mistake that I didn't simply change tan x = t and dx = dt/t^2 + 1, instead I expressed tan x as sin x / cos x and divided the denominator with cos^2 x and it turned into crazy powers of t. Your method works great, thanks.
 

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