Trigonometric Substitution Proof

  • Thread starter LHC
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  • #1
LHC
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The question is:

Use [tex]x = \tan \theta , \frac{-\pi}{2} < \theta < \frac{\pi}{2} [/tex] to show that:

[tex] \int_{0}^{1} \frac{x^3}{\sqrt{x^2+1}} dx =\int_{0}^{\frac{\pi}{4}} \tan^3 \theta \sec \theta d\theta[/tex]

Using that substitution, I got it down to:

[tex]\int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sqrt{\tan^2 \theta+1}} = \int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sec \theta}[/tex]

I have no clue how this is going to get to the answer. Could someone please help? Thanks.
 

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
476
1
What about the dx term?

[tex]
dx=d(\tan\theta)=??
[/tex]
 
  • #3
Dick
Science Advisor
Homework Helper
26,263
619
You are forgetting the dx part again. That's why your expression differs from what you are supposed to show. When you get it right, try expressing the integrand in terms of sin and cos.
 
  • #4
LHC
24
0
Ah, NOW I get it. Thanks to everyone for your help!
 

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