Solve Integral Problem: Evaluate $\displaystyle\int^{1}_{0}{\sqrt{x^2+1}}$

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SUMMARY

The integral $\displaystyle\int^{1}_{0}{\sqrt{x^2+1}}$ can be evaluated using trigonometric substitution, specifically setting $x = \tan{\theta}$, which transforms the integral into $\int^{\frac{\pi}{4}}_{0}{\sec^3{\theta}}\,d\theta$. The solution involves recognizing that $\sec^3{\theta}$ can be decomposed into $\sec{\theta}\tan^2{\theta} + \sec{\theta}$, leading to a more manageable integral. The discussion highlights the importance of differentiation techniques, particularly the derivative of $\sec{\theta}\tan{\theta}$, to progress further in solving the integral.

PREREQUISITES
  • Trigonometric substitution techniques
  • Understanding of integrals involving secant and tangent functions
  • Knowledge of differentiation rules, particularly for secant and tangent
  • Familiarity with integral limits and transformations
NEXT STEPS
  • Study the integration of $\sec^3{\theta}$ and its applications
  • Learn about trigonometric identities and their use in integration
  • Explore advanced techniques in integral calculus, such as integration by parts
  • Practice solving integrals involving trigonometric substitutions
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Students studying calculus, particularly those focusing on integral calculus and trigonometric functions, as well as educators looking for effective teaching methods for integral evaluation techniques.

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Homework Statement


[tex]Evaluate $\displaystyle\int^{1}_{0}{\sqrt{x^2+1}}$[/tex]


Homework Equations





The Attempt at a Solution


[tex]By trigonometric substitution: $x = \tan{\theta} \rightarrow dx = \sec^2{\theta}\,d\theta$<br /> \[\int^{\frac{\pi}{4}}_{0}{\sec^2{x}\sqrt{\tan^2{\theta}+1}}\,d\theta = \int^{\frac{\pi}{4}}_{0}{\sec^3{\theta}}\,d\theta\]<br /> \[= \int^{\frac{\pi}{4}}_{0}{\sec{\theta}\tan^2{\theta}+\sec{\theta}}\,d\theta\][/tex]

This is where I get stuck
 
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imranq said:
[tex]\int^{\frac{\pi}{4}}_{0}{\sec^2{\theta}\sqrt{\tan^2{\theta}+1}}\,d\theta = \int^{\frac{\pi}{4}}_{0}{\sec^3{\theta}}\,d\theta\]<br /> \[= \int^{\frac{\pi}{4}}_{0}{\sec{\theta}\tan^2{\theta}+\sec{\theta}}\,d\theta\][/tex]

This is where I get stuck

Hi imranq! :smile:

(have a theta: θ and a squared: ² and a cubed: ³ :smile:)

Hint: (d/dθ)(secθ tanθ) = … ? :wink:
 

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