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Given a continuous function $f:[0,1] \to \mathbb {R}$

  1. Prove that $$\int^{\pi/2}_0 f(\cos x)dx = \int^{\pi/2}_0 f(\sin x) dx$$

  2. Calculate $$I= \int^{\pi/2}_0 \frac{\sin^2 x + \sin x}{1 + \sin x + \cos x} dx$$

The 1st was solved by letting $g:[0,1] \to \mathbb{R}$ such that $g(x) = f(\sin x)$ and proving that $\int^{\pi/2}_0 g(x) dx = \int^{\pi/2}_0 g(\pi /2 -x) dx$ using a substitution.

My question: Can I calculate $I$ using the first lemma? I know that by using other methods the integral turns out to be $I = \bigg [(x-\sin x - \cos x)/2 \bigg |^{\pi/2}_0 \bigg ] = \pi/4 $

Finally, note that $I= \int^{\pi/2}_0 \frac{\sin^2 x + \sin x}{1 + \sin x + \cos x} dx =\int^{\pi/2}_0 \frac{\cos^2 x + \cos x}{1 + \cos x + \sin x} dx $

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    $\begingroup$ "My question: Can I calculate I using the first lemma?" Yes. $\endgroup$ – Did Feb 5 '15 at 16:41
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Use the substitution $u=\frac{\pi}{2}-x$ (first lemma) to conclude that our second integral is $$\frac{1}{2}\int_0^{\pi/2}\frac{\sin^2 x+\sin x+\cos^2 x+\cos x}{1+\sin x+\cos x}\,dx.$$ But $\sin^2 x+\cos^2 x=1$, so we are integrating $1$!.

Remark: I don't really think of it as substitution, it is symmetry that does it.

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