- #1
LHC
- 24
- 0
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}
I have no clue how this is going to get to the answer. Could someone please help? Thanks.
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}
I have no clue how this is going to get to the answer. Could someone please help? Thanks.