Daily math problems for September 7, 2026 — Trigonometry, Integrals, Matrices & more

$2{\cos }^{2}x+\cos x-1=0\Rightarrow (2\cos x-1)(\cos x+1)=0$. Roots: $\cos x=\frac{1}{2}$ ($x=\frac{\pi }{3},\frac{5\pi }{3}$) and $\cos x=-1$ ($x=\pi$). Three solutions.

Let $u=\sin x$, $du=\cos xdx$. The integral becomes ${\int }_0^1{u}^{3}du={\left[{\displaystyle \frac{u^4}{4}}\right]}_0^1={\displaystyle \frac{1}{4}}$.

$(AB{)}_{11}=1\cdot 3+2\cdot 1=5$; $(AB{)}_{12}=1\cdot 0+2\cdot 4=8$; $(AB{)}_{21}=0\cdot 3+1\cdot 1=1$; $(AB{)}_{22}=0\cdot 0+1\cdot 4=4$. So $AB=\left(\begin{array}{cc}5& 8\\ 1& 4\end{array}\right)$.

${\displaystyle \frac{x^2-1}{x-1}}=x+1\to 2$ as $x\to 1$. For continuity, $a=2$.

Expand along the first column: $1\cdot (1\cdot 0-4\cdot 6)-0+5\cdot (2\cdot 4-3\cdot 1)=-24+5\cdot 5=1$.

$\bullet$ Classify the first two terms by parity. $x^3\cos x$ is odd $\times$ even $=$ odd, and $x^5{\sin }^{2}x$ is odd $\times$ even $=$ odd. Over the symmetric interval $[-\pi ,\pi ]$:

$${\int }_{-\pi }^{\pi }{x}^3\cos xdx=0$$

$${\int }_{-\pi }^{\pi }{x}^5{\sin }^{2}xdx=0$$

$\bullet$ The third term vanishes at the endpoints:

$${\int }_{-\pi }^{\pi }\cos xdx=[\sin x{]}_{-\pi }^{\pi }=0$$

$\bullet$ Decode the constants:

$${\log }_{3}81=4,2!=2$$

$\bullet$ Combine:

$$0+4-2=2$$

$u=x^2+1\Rightarrow du=2xdx$. When $x=0$, $u=1$; when $x=1$, $u=2$. ${\int }_1^2{u}^{3}du={\displaystyle \frac{16-1}{4}}={\displaystyle \frac{15}{4}}$.

Folosind schimbarea de bază, ${\log }_k(k+1)={\displaystyle \frac{\ln (k+1)}{\ln k}}$. Produsul telescopează: ${\displaystyle \frac{\ln 3}{\ln 2}}\cdot {\displaystyle \frac{\ln 4}{\ln 3}}\cdots {\displaystyle \frac{\ln 256}{\ln 255}}={\displaystyle \frac{\ln 256}{\ln 2}}={\log }_{2}256=8$.

${\displaystyle \frac{1}{n}}\sum (x_i+5)={\displaystyle \frac{1}{n}}\sum x_i+5=\bar{x}+5$.

After both moves, the value is $0.8\cdot 1.2=0.96$ — a $4\mathrm{\%}$ net loss.