Puzzles
An integral
What is
$$\int_0^{\frac\pi2}\frac1{1+\tan^a(x)}\,dx?$$
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Surprisingly, the value of the integral does not depend on \(a\). To show this, first use \(\tan(x)=\frac{\sin(x)}{\cos(x)}\) on both integrals:
$$I_1=\int_0^{\frac\pi2}\frac1{1+\tan^a(x)}\,dx
$$$$=\int_0^{\frac\pi2}\frac{\cos^a(x)}{\cos^a(x)+\sin^a(x)}\,dx$$
$$I_2=\int_0^{\frac\pi2}\frac1{1+\cot^a(x)}\,dx
$$$$=\int_0^{\frac\pi2}\frac{\sin^a(x)}{\cos^a(x)+\sin^a(x)}\,dx$$
Now consider \(I_1+I_2\):
$$I_1+I_2 = \int_0^{\frac\pi2}\frac{\cos^a(x)+\sin^a(x)}{\cos^a(x)+\sin^a(x)}\,dx$$$$=\int_0^{\frac\pi2}1\,dx$$$$=\frac\pi2$$
By substituting \(u=\tfrac\pi4-x\) into \(I_2\), it can be shown that \(I_1 = I_2\).
Therefore \(I_1=\tfrac\pi4\).
Extension
What is
$$\int_0^{\frac\pi2}\frac1{1-\tan^a(x)}\,dx?$$
Find them all
Find all continuous positive functions, \(f\) on \([0,1]\) such that:
$$\int_0^1 f(x) dx=1\\
\mathrm{and }\int_0^1 xf(x) dx=\alpha\\
\mathrm{and }\int_0^1 x^2f(x) dx=\alpha^2$$
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$$0=\alpha^2-2\alpha^2+\alpha^2\\
=\int_0^1 x^2f(x) dx-2\int_0^1 \alpha xf(x) dx+\int_0^1 \alpha^2f(x)dx\\
=\int_0^1 (x^2-2\alpha x+\alpha^2)f(x)dx\\
=\int_0^1 (x-\alpha)^2 f(x)dx$$
\(f(x)\) and \((x-\alpha)^2\) are both positive so this is only possible if one of them if always zero.
But \((x-\alpha)^2\) is only zero when \(x=\alpha\) and \(\int_0^1 f(x) dx=1\) so \(f(x)\) cannot always be zero. Therefore no such function exists.
Extension
Find all continuous positive functions, \(f\) on \([0,1]\) such that:
$$\int_0^1 f(x) dx=1\\
\int_0^1 xf(x) dx=1\\$$
Integrals
$$\int_0^1 1 dx = 1$$
Find \(a_1\) such that:
$$\int_0^{a_1} x dx = 1$$
Find \(a_2\) such that:
$$\int_0^{a_2} x^2 dx = 1$$
Find \(a_n\) such that (for \(n>0\)):
$$\int_0^{a_n} x^n dx = 1$$
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$$1=\int_0^{a_1} x dx=\frac{a_1^2}{2}$$
So, \(a_1=\sqrt{2}\).
$$1=\int_0^{a_2} x^2 dx=\frac{a_2^3}{3}$$
So, \(a_2=3^{\frac{1}{3}}\).
$$1=\int_0^{a_n} x^n dx=\frac{a_n^{n+1}}{n+1}$$
So, \(a_n={(n+1)}^{\frac{1}{n+1}}\).
Extension
Find \(b_n\) such that (for \(n>1\)):
$$\int_{b_n}^{\infty} x^{-n} dx = 1$$
