Evaluate \int \sqrt{\frac{1 + x^2}{x}} dx using trig substitution (i.e. x = \tan (\theta, dx =...


Evaluate {eq}\int \sqrt{\frac{1 + x^2}{x}} dx {/eq} using trig substitution (i.e. {eq}x = \tan (\theta, dx = sec^2(\theta)) {/eq}

Let {eq}\displaystyle x = \tan (\theta) {/eq}

Then we have the following results:

  • {eq}\displaystyle \sqrt(1+x^2)=\sqrt(1+tan^2\theta)=\sqrt(sec^2\theta)=sec\theta \\ {/eq}
  • {eq}\displaystyle \frac{1}{x}=\frac{1}{tan\theta}=cot\theta \\ {/eq}
  • {eq}\displaystyle \frac{\sqrt(1+x^2)}{x}=\frac{sec\theta}{tan\theta}=cosec\theta {/eq}

Answer and Explanation: 1

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{eq}\displaystyle \int \frac{\sqrt{1+x^2}}{x}\:dx \\ {/eq}

Let {eq}\displaystyle x = \tan (\theta) ; dx = sec^2(\theta) d\theta \\ {/eq}


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Indefinite Integrals as Anti Derivatives


Chapter 12 / Lesson 11

Indefinite integrals, an integral of the integrand which does not have upper or lower limits, can be used to identify individual points at specific times. Learn more about the fundamental theorem, use of antiderivatives, and indefinite integrals through examples in this lesson.

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