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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.