Daily Integral 11 February 2026

Daily Integral 11/02/26

Not the hardest integrals, but still do hard integrals yet. Got into this because all my friends are doing it.

Beginner Integral

$$\begin{gathered} I = \int_0^\frac{\pi}{3} sec\ x(sec\ x+ tan\ x)\ dx \\ \ \\ \text{Expand the brackets}\\ =\int_0^\frac{\pi}{3} sec^2\ x + sec\ x\cdot tan\ x\ dx \\ \ \\ \text{Separate the integrands}\\ =\int_0^\frac{\pi}{3} sec^2\ x\ dx + \int_0^\frac{\pi}{3}sec\ x\cdot tan\ x\ dx \\ \ \\ \text{Solve for $\int_0^\frac{\pi}{3} sec^2\ x\ dx$}\\ =[tan\ x]_0^\frac{\pi}{3} + \int_0^\frac{\pi}{3}sec\ x\cdot tan\ x\ dx \\ =\sqrt{3} + \int_0^\frac{\pi}{3}sec\ x\cdot tan\ x\ dx \\ \ \\ \text{Combine $sec\ x\ \cdot tan\ x$}\\ =\sqrt{3} + \int_0^\frac{\pi}{3}\frac{sin\ x}{cos^2\ x}\ dx \\ \ \\ \text{let $u = cos\ x,\ \frac{du}{dx} = -sin\ x$} \\ =\sqrt{3} + \int_1^\frac{1}{2}-\frac{1}{u^2}\ du \\ =\sqrt{3} + [u^{-1}]_1^\frac{1}{2} \\ =\boxed{\sqrt{3} + 1} \end{gathered}$$

I was stuck for a bit due to failing to remember $\frac{d}{dx}(cos\ x) = -sin\ x$. I thought it was $\frac{d}{dx}(cos\ x) = sin\ x$

Easy Integral

$$\begin{gathered} I = \int_\frac{\pi}{3}^\frac{\pi}{6} \left(\frac{2cos(2x)}{sin(2x)} \left[\left(\frac{cos(2x)}{cos\ x - sin\ x}\right) ^2 - sin(2x)\right] - cot\ x\right)\ dx \\ \ \\ \text{Simplify inner-inner brackets} \\ \left(\frac{cos(2x)}{cos\ x - sin\ x}\right) ^2 \\ =\frac{cos^2(2x)}{cos^2\ x + sin^2\ x - 2\ cos\ x\ sin\ x} \\ =\frac{cos^2(2x)}{1 - 2\ cos\ x\ sin\ x} \\ =\frac{cos^2(2x)}{1 - sin(2x)} \\ \ \\ \text{Simplify inner brackets} \\ \left(\frac{cos^2(2x)}{1 - sin(2x)} - sin(2x)\right)^2 \\ =\left(\frac{cos^2(2x)}{1 - sin(2x)} - \frac{sin(2x)(1-sin(2x)}{1-sin(2x)}\right)^2 \\ =\left(\frac{cos^2(2x) - sin(2x)(1-sin(2x)}{1 - sin(2x)}\right)^2 \\ =\left(\frac{cos^2(2x) - sin(2x) + sin^2(2x)}{1 - sin(2x)}\right)^2 \\ =\left(\frac{1 - sin(2x)}{1 - sin(2x)}\right)^2 \\ =1 \\ \ \\ \text{Simplify outer brackets} \\ \frac{2cos(2x)}{sin(2x)} - cot\ x \\ =\frac{2cos(2x)}{sin(2x)} - \frac{cos\ x}{sin\ x} \\ =\frac{2\ cos^2\ x - 2\ sin^2\ x - 2\ cos^2\ x}{2\ sin\ x\ cos\ x} \\ =-\frac{sin^2\ x}{sin\ x\ cos\ x} \\ =-\frac{sin\ x}{ cos\ x} \\ =-tan\ x \\ \ \\ \text{Sub into original integrand} \\ \int_\frac{\pi}{3}^\frac{\pi}{6} -tan\ x\ dx \\ \ \\ \text{let $u = cos\ x\ \frac{du}{dx} = -sin\ x$} \\ I =-\int_\frac{1}{2}^\frac{\sqrt3}{2}\frac{1}{u}\ du \\ =-[ln|u|]_\frac{1}{2}^\frac{\sqrt3}{2} \\ =-(ln(\frac{\sqrt3}{2})-ln(\frac{1}{2})) \\ =\boxed{-ln(\sqrt3)} \end{gathered}$$

Medium Integral

$$\begin{gathered} I = \int_0^2 \frac{4|x-2||x|}{(x+1)(x+2)^2}\ dx \\ \ \\ \text{Since the integration is from 0 to 2, we can observe that} \\ \text{$|x-2| = 2-x$ and $|x| = x$} \\ \therefore I = \int_0^2 \frac{4x(2-x)}{(x+1)(x+2)^2}\ dx \\ \ \\ \text{Dissolve the fraction by partial fraction} \\ \frac{4x(2-x)}{(x+1)(x+2)^2}\ = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}\\ 8x - 4x^2 = A(x+2)^2 + B(x+1)(x+2) + C(x+1) \\ \ \\ \text{Set $x = -2$} \\ \therefore -32 = -C \\ \ \\ \text{Set $x = -1$} \\ \therefore -12 = A \\ \ \\ \text{Set $x = -2$} \\ \begin{aligned} 0 &= 4A+2B+C \\ &=-48+2B+32 \\ &=-16+2B \\ \end{aligned} \\ \therefore B=8 \ \\ \frac{4x(2-x)}{(x+1)(x+2)^2}\ = -\frac{12}{x + 1} + \frac{8}{x + 2} + \frac{32}{(x + 2)^2}\\ \ \\ \text{Sub into original integrand} \\ \left[-12\ ln\ |x+1| + 8\ ln(x+2) - \frac{32}{x+2}\right]^2_0 \\ =(-12\ ln\ 3 + 8\ ln\ 4 - 8) - (8\ ln\ 2 + 16) \\ =-12\ ln\ 3 + 16\ ln\ 2 - 8\ ln\ 2 - 8 + 16 \\ =\boxed{-12\ ln\ 3 + 8\ ln\ 2 + 8} \\ \end{gathered}$$