Daily math problems for August 1, 2026 — Calculus, Trigonometry, Analytic Geometry & more

$\sin x\ge 0$ on $[0,\pi \mathrm{/}2]$, so ${\sin }^{n}x\ge 0$ and the integral is $\ge 0$ (in fact $>0$ for finite $n$).

$\sin \text{\hspace{0.17em}⁣}(2\pi -{\displaystyle \frac{\pi }{6}})=-\sin {\displaystyle \frac{\pi }{6}}=-{\displaystyle \frac{1}{2}}$.

Base $=6$, height $=4$. Area $={\displaystyle \frac{1}{2}}\cdot 6\cdot 4=12$.

$R(\theta )R(-\theta )=I_2$, so $R(\theta {)}^{-1}=R(-\theta )$.

Left branch ($x<0$): $f(x)=x+1\in (-\mathrm{\infty },1)$, strictly increasing. Right branch ($x\ge 0$): $f(x)=2x+1\in [1,+\mathrm{\infty })$, strictly increasing.

The two images $(-\mathrm{\infty },1)$ and $[1,+\mathrm{\infty })$ are disjoint and together cover all of $\mathbb{R}$, so $f$ is surjective. Each branch is strictly increasing, and the ranges don't overlap, so no two distinct $x$-values give the same $f(x)$ — $f$ is injective. Hence $f$ is bijective.

(Note: $f(-1)=0$ and $f(0)=1$, so distractor A is factually false.)

$\bullet$ Differentiate both sides implicitly (product rule on $xy$):

$$2x+y+x{y}^{\mathrm{\prime }}+2y{y}^{\mathrm{\prime }}=0$$

$\bullet$ Substitute $(x,y)=(1,1)$:

$$2+1+y^{\mathrm{\prime }}+2y^{\mathrm{\prime }}=0$$

$\bullet$ Solve for $y^{\mathrm{\prime }}$:

$$3y^{\mathrm{\prime }}=-3\Rightarrow y^{\mathrm{\prime }}=-1$$

$\cos u=-\frac{1}{2}$ on $[0,4\pi )$ at $u=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}$. Dividing by $2$: $x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$. Four solutions.

Substitute $x=\tan \theta$. Then $dx={\sec }^2\theta d\theta$ and $1+x^2={\sec }^2\theta$, so the ${\sec }^2\theta$ on top and bottom cancel cleanly. The bounds $0\to 1$ map to $\theta :0\to \pi \mathrm{/}4$:

$$I={\int }_0^{\pi \mathrm{/}4}\ln (1+\tan \theta )d\theta \mathrm{.}$$

Reflect by $\varphi =\pi \mathrm{/}4-\theta$. The angle-difference formula gives $\tan (\pi \mathrm{/}4-\varphi )={\displaystyle \frac{1-\tan \varphi }{1+\tan \varphi }}$, so

$$1+\tan \theta =1+\frac{1-\tan \varphi }{1+\tan \varphi }=\frac{2}{1+\tan \varphi }\mathrm{.}$$

Taking logs: $\ln (1+\tan \theta )=\ln 2-\ln (1+\tan \varphi )$.

Pair the integral with its reflected twin. Renaming the dummy variable $\varphi \to \theta$ (the bounds are unchanged because the substitution is a reflection of $[0,\pi \mathrm{/}4]$ onto itself):

$$I={\int }_0^{\pi \mathrm{/}4}[\ln 2-\ln (1+\tan \theta )]d\theta =\frac{\pi }{4}\ln 2-I\mathrm{.}$$

Solve. $2I={\displaystyle \frac{\pi }{4}}\ln 2$, so

$$\boxed{{\displaystyle I=\frac{\pi }{8}\ln 2\mathrm{.}}}$$

The reflection trick — pairing $f(\theta )$ with $f(\pi \mathrm{/}4-\theta )$ — works whenever the integrand simplifies under that reflection. Worth keeping in your toolbox alongside the half-angle and Weierstrass substitutions.

Direct substitution gives $\frac{0}{0}$. Differentiate top and bottom: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{2x}}}$ — still $\frac{0}{0}$. Once more: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x}{2}}={\displaystyle \frac{1}{2}}}$.

$n\ln \text{\hspace{0.17em}⁣}(1+\frac{1}{n^2})\sim n\cdot \frac{1}{n^2}=\frac{1}{n}\to 0$, so the limit equals $e^0=1$.