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 $

  • 1
    $\begingroup$ "My question: Can I calculate I using the first lemma?" Yes. $\endgroup$ – Did Feb 5 '15 at 16:41

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.