IMO Practice Test — Integrals
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
$\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx=$
$\pi^2/4$
$\pi/2$
$\pi^2/2$
$\pi$
Explanation: King's rule: $I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx$. Adding: $2I=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx$. Sub $t=\cos x$: $2I=\pi\int_{-1}^1\frac{dt}{1+t^2}=\pi\cdot\pi/2$. $I=\pi^2/4$.
Question 2 of 6
hard
$\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx=$
$\frac{\pi\ln 2}{8}$
$\frac{\pi}{4}\ln 2$
$\frac{\pi}{8}$
$\ln 2$
Explanation: Classic result: $\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi\ln 2}{8}$ (proved via $x=\tan\theta$ substitution and King's property).
Question 3 of 6
hard
$\int\frac{\sin x}{\sin x + \cos x}\,dx=$
$\frac{x}{2}+\frac{1}{2}\ln|\sin x+\cos x|+C$
$\frac{x}{2}-\frac{1}{2}\ln|\sin x+\cos x|+C$
$\ln|\sin x+\cos x|+C$
$\frac{x}{2}+C$
Explanation: Write $\sin x = \frac{1}{2}[(\sin x+\cos x)+(\sin x-\cos x)]$. Then split integral and integrate each part.
Question 4 of 6
hard
$\int_0^{\pi/2}\frac{dx}{1+\tan^n x}$ for any positive integer $n$:
Depends on $n$
$\pi/2$
$\pi/4$
$0$
Explanation: $I=\int_0^{\pi/2}\frac{\cos^n x}{\sin^n x+\cos^n x}dx$. By King's property ($x\to\pi/2-x$): $I=\int_0^{\pi/2}\frac{\sin^n x}{\cos^n x+\sin^n x}dx$. Adding: $2I=\pi/2$, $I=\pi/4$ for all $n$.
Question 5 of 6
medium
$\int_0^1[x]\,dx$ where $[x]$ is the greatest integer function:
$0$
$1$
$1/2$
$1/4$
Explanation: On $[0,1)$, $[x]=0$. At $x=1$, $[x]=1$ but it's a single point (measure zero). So $\int_0^1[x]dx=0$.
Question 6 of 6
hard
If $I_n=\int_0^{\pi/4}\tan^n x\,dx$, then $I_n+I_{n-2}=$
$\frac{1}{n-1}$
$\frac{1}{n+1}$
$\frac{1}{n}$
$\frac{2}{n-1}$
Explanation: $I_n+I_{n-2}=\int_0^{\pi/4}\tan^{n-2}x(\tan^2x+1)dx=\int_0^{\pi/4}\tan^{n-2}x\sec^2x\,dx=[\tan^{n-1}x/(n-1)]_0^{\pi/4}=\frac{1}{n-1}$.