# Trigonometric Substitution Proof

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:

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

jamesrc
Gold Member

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

Dick