Struggling with a Deceptively Difficult Integral

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Homework Statement


It looks deceptively easy, but I can't seem to get it...
\int^{\frac{\pi}{4}}_{0}{\\sec^4{x}\tan^4{x}}\,dx

Homework Equations



\sec^2{x} = \tan^2{x} +1

The Attempt at a Solution


I've tried, but they all end up as trigonometric polynomials
 
Last edited:
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It IS easy. Substitute u=tan(x).
 
ah shoot! Thanks, now I got it
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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