Trig. Indefinite Integral

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SUMMARY

The integral evaluation of $\displaystyle \int\sqrt\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx$ was discussed, with attempts made using the substitutions $\displaystyle \tan x = \frac{2\tan \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$, $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$, and $\displaystyle \sin x = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$. However, no standard substitution form was found. The Risch Algorithm was suggested as a potential method to determine if the integral can be expressed in terms of elementary functions.

PREREQUISITES
  • Understanding of trigonometric identities and substitutions
  • Familiarity with integral calculus and indefinite integrals
  • Knowledge of the Risch Algorithm for determining elementary antiderivatives
  • Proficiency in manipulating expressions involving $\tan$, $\sin$, and $\cos$ functions
NEXT STEPS
  • Research the Risch Algorithm and its application in integral calculus
  • Study advanced techniques for evaluating indefinite integrals
  • Explore trigonometric substitutions in integral calculus
  • Learn about special functions and their relationship to elementary functions
USEFUL FOR

Mathematicians, calculus students, and anyone interested in advanced integral evaluation techniques, particularly those involving trigonometric functions.

juantheron
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Evaluation of $\displaystyle \int\sqrt\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx$

I have Tried The Given Integral Using $\displaystyle \tan x = \frac{2\tan \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ and $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ and $\displaystyle \sin x = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$

but Could not find anything in standard Substution form

Help me

Thanks
 
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jacks said:
Evaluation of $\displaystyle \int\sqrt\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx$

I have Tried The Given Integral Using $\displaystyle \tan x = \frac{2\tan \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ and $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ and $\displaystyle \sin x = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$

but Could not find anything in standard Substution form

Help me

Thanks

Hi jacks, :)

Just out of curiosity where did you found this integral? Just a guess but this integral might be expressed through elementary functions. I am not quite sure, but as far as I know one can use the Risch Algorithm to determine whether a function has an elementary antiderivative.

[graph]yi8y5ym7yx[/graph]
 

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