Trigonometric Substitution Proof

1. Jul 22, 2008

LHC

The question is:

Use $$x = \tan \theta , \frac{-\pi}{2} < \theta < \frac{\pi}{2}$$ to show that:

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

Using that substitution, I got it down to:

$$\int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sqrt{\tan^2 \theta+1}} = \int_{0}^{\frac{\pi}{4}} \frac{\tan^3 \theta}{\sec \theta}$$

2. Jul 22, 2008

jamesrc

$$dx=d(\tan\theta)=??$$

3. Jul 22, 2008

Dick

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. Jul 23, 2008

LHC

Ah, NOW I get it. Thanks to everyone for your help!