Chebyshev-Gauss quadrature with $\tan(x)$

Question

With Chebyshev-Gauss quadrature, solve $\int_0^{\pi/4}x\cdot tan^2(x)$, for $n=3$.

Needs first to determine the change in the integral, to change the limits of integrals and then reduce in form integrate $$\int_{-1}^1{\dfrac{1}{(1-x^2)^{1/2}}}$$ http://upload.wikimedia.org/math/3/7/3/3739c7537ace93cb3bc05e3957a44ff3.png

Thanks a lot, this is really important for me.

Greetings, Tanya

This is for exam in Faculty of Sciences. I cant change tan^2(x) to (1-x^2)^-1/2 and also change limits [0,Pi/4]->[-1,1] — , May 28 '13 at 11:00
I think you should probably be able to do this if you make some sort of inverse trig substitution for $x$, possibly $x = \arcsin(t)$ or $x = \arccos(t)$ or something similar. — in_mathematica_we_trust, May 28 '13 at 11:04
For some basic information about writing math at this site see e.g. here, here, here and here. — Julian Kuelshammer, May 28 '13 at 11:08

score 0 · Accepted Answer · answered May 28 '13 at 11:32

Hint: make the substitution $$4x - \frac{\pi}{2} = \arcsin(t).$$

You will need to simplify some horrific looking trig. functions in the integrand but this substitution should give you the correct limits of integration and a factor of $\sqrt{1 - t^2}$ in the denominator.

Chebyshev-Gauss quadrature with $\tan(x)$

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