How to solve these integrals? sqrt(a^2 + x^2) & sqrt(2 + x^2)?

  • Context: Undergrad 
  • Thread starter Thread starter srk999
  • Start date Start date
  • Tags Tags
    Integrals
Click For Summary
SUMMARY

This discussion focuses on solving the integrals of the forms sqrt(a^2 + x^2) and sqrt(2 + x^2) using trigonometric substitutions. The recommended substitution for the first integral is x = a tan(θ), leading to the simplification of the integral to |a sec(θ)|. The second integral is directly analogous to the first, with the specific case of a = sqrt(2). Both integrals can be approached without a calculator, relying solely on sine and cosine functions.

PREREQUISITES
  • Understanding of trigonometric identities, particularly secant and tangent functions.
  • Familiarity with hyperbolic functions, specifically sinh(θ).
  • Knowledge of integral calculus and substitution methods.
  • Basic algebraic manipulation skills for simplifying expressions.
NEXT STEPS
  • Study trigonometric substitution techniques in integral calculus.
  • Learn about hyperbolic functions and their applications in calculus.
  • Practice solving integrals involving sqrt(a^2 + x^2) using various substitution methods.
  • Explore numerical integration techniques for evaluating integrals without a calculator.
USEFUL FOR

Students preparing for calculus exams, particularly those focusing on integral calculus, and anyone looking to enhance their problem-solving skills in mathematics.

srk999
Messages
6
Reaction score
0
How to solve these integrals, such as- sqrt(a^2 + x^2) & sqrt(2 + x^2)

Please be as descriptive and simple as possible.

Please use only sin and cos if possible We are not allowed a calculator in the exam and will have to find numerical values.
 
Physics news on Phys.org
For the first: try the substitution [tex]x = a \sinh(\theta) = a\frac{e^{\theta}-e^{-\theta}}{2}[/tex]
 
Or use a trig substitution: [itex]x= a tan(\theta)[/itex]. Then [itex]a^2+ x^2= a^2+ a^2tan^2(\theta)[/itex][itex]= a^2(1+ tan^2(\theta))= a^2 sec^2(\theta)[/itex]. [itex]\sqrt{a^2+ x^2}= |a sec(x)|.<br /> <br /> And, of course, [itex]dx= a(tan(\theta))' d\theta= a sec^2(\theta)d\theta[/itex]<br /> <br /> I hope you recognize that the second problem, with [itex]\sqrt{2+ x^2}[/itex], is exactly the same as the first problem with [itex]a= \sqrt{2}[/itiex].[/itex][/itex]
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K