Trigonometric Substitution Proof

LHC
Messages
24
Reaction score
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.
 
Physics news on Phys.org
What about the dx term?

<br /> dx=d(\tan\theta)=??<br />
 
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.
 
Ah, NOW I get it. Thanks to everyone for your help!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top