Daily Integral 15 February 2026
Pretty nice integrals in general, nothing too hard. The medium got me stuck because I didn’t recognise the integral is simply a u-sub with e e e cot x \cot x cot x
∫ 0 2 x 2 − 9 x − 3 d x = ∫ 0 2 x + 3 d x = [ 1 2 x 2 + 3 x ] 0 2 = 8 \begin{gathered}
\int_0^2 \frac{x^2-9}{x-3}\ dx \\
= \int_0^2 x+3\ dx \\
= \left[\frac{1}{2}x^2 + 3x\right]^2_0 \\
=\boxed 8
\end{gathered}
∫ 0 2 x − 3 x 2 − 9 d x = ∫ 0 2 x + 3 d x = [ 2 1 x 2 + 3 x ] 0 2 = 8
I = ∫ 1 e ( ln x + 1 ) ( ln 2 x − ln x + 1 ) x d x = ∫ 1 e ln 3 x − ln 2 x + ln x + ln 2 x − ln x + 1 x d x = ∫ 1 e ln 3 x + 1 x d x = ∫ 1 e ln 3 x x d x + ∫ 1 e 1 x d x = ∫ 1 e ln 3 x x d x + [ ln ∣ x ∣ ] 1 e = ∫ 1 e ln 3 x x d x + 1 Let u = ln x , d u = x − 1 d x I = ∫ 0 1 u 3 d u + 1 = [ 1 4 u 4 ] 0 1 + 1 = 5 4 \begin{gathered}
I = \int_1^e \frac{(\ln x + 1)(\ln^2x - \ln x + 1)}{x}\ dx \\
= \int_1^e \frac{\ln^3 x - \ln^2 x + \ln x + \ln^2 x - \ln x + 1}{x}\ dx \\
= \int_1^e \frac{\ln^3 x + 1}{x}\ dx \\
= \int_1^e \frac{\ln^3 x}{x}\ dx + \int_1^e \frac{1}{x}\ dx \\
= \int_1^e \frac{\ln^3 x}{x}\ dx + [\ln|x|]^e_1 \\
= \int_1^e \frac{\ln^3 x}{x}\ dx + 1 \\
\text{Let $u = \ln x$, $du = x^{-1}\ dx$} \\
I = \int_0^1 u^3\ du + 1 \\
= \left[\frac{1}{4} u^4\right]^1_0 + 1 \\
= \boxed{\frac{5}{4}}
\end{gathered}
I = ∫ 1 e x ( ln x + 1 ) ( ln 2 x − ln x + 1 ) d x = ∫ 1 e x ln 3 x − ln 2 x + ln x + ln 2 x − ln x + 1 d x = ∫ 1 e x ln 3 x + 1 d x = ∫ 1 e x ln 3 x d x + ∫ 1 e x 1 d x = ∫ 1 e x ln 3 x d x + [ ln ∣ x ∣ ] 1 e = ∫ 1 e x ln 3 x d x + 1 Let u = ln x , d u = x − 1 d x I = ∫ 0 1 u 3 d u + 1 = [ 4 1 u 4 ] 0 1 + 1 = 4 5
I = ∫ π 4 π 2 e cos x ( cos 3 x − cos x − 1 1 − cos 2 x d x = ∫ π 4 π 2 e cos x ( cos x ( 1 − sin 2 x ) − cos x − 1 ) sin 2 x = ∫ π 4 π 2 e cos x ( cos x − cos x sin 2 x − cos x − 1 ) sin 2 x = ∫ π 4 π 2 e cos x ( − cos x sin 2 x − 1 ) sin 2 x = ∫ π 4 π 2 − e cos x cos x sin 2 x − e cos x sin 2 x = ∫ π 4 π 2 − e cos x cos x − e cos x csc 2 x d x Let u = e cos x cot x d u = − sin x e cos x cot x − e cos x csc 2 x d x = − cos x e cos x − e cos x csc 2 x d x I = [ e cos x cot x ] π 4 π 2 = − e 2 2 \begin{gathered}
I = \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{e^{\cos x}(\cos^3 x - \cos x - 1}{1 - \cos^2 x}\ dx \\
= \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{e^{\cos x}(\cos x(1 - \sin^2 x) - \cos x - 1)}{\sin^2 x} \\
= \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{e^{\cos x}(\cos x - \cos x \sin^2 x - \cos x - 1)}{\sin^2 x} \\
= \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{e^{\cos x}(-\cos x\sin^2 x - 1)}{\sin^2 x} \\
= \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{-e^{\cos x}\cos x\sin^2 x - e^{\cos x}}{\sin^2 x} \\
= \int_\frac{\pi}{4}^\frac{\pi}{2} -e^{\cos x}\cos x - e^{\cos x}\csc^2 x\ dx \\
\ \\
\text{Let $u = e^{\cos x}\cot x$} \\
\begin{aligned}
du &= -\sin x e^{\cos x}\cot x - e^{\cos x}\csc^2 x\ dx \\
&= -\cos x e^{\cos x} - e^{\cos x}\csc^2 x\ dx\\
\end{aligned} \\
I = \left[e^{\cos x}\cot x\right]_\frac{\pi}{4}^\frac{\pi}{2} \\
= \boxed{-e^\frac{\sqrt 2}{2}}
\end{gathered}
I = ∫ 4 π 2 π 1 − cos 2 x e c o s x ( cos 3 x − cos x − 1 d x = ∫ 4 π 2 π sin 2 x e c o s x ( cos x ( 1 − sin 2 x ) − cos x − 1 ) = ∫ 4 π 2 π sin 2 x e c o s x ( cos x − cos x sin 2 x − cos x − 1 ) = ∫ 4 π 2 π sin 2 x e c o s x ( − cos x sin 2 x − 1 ) = ∫ 4 π 2 π sin 2 x − e c o s x cos x sin 2 x − e c o s x = ∫ 4 π 2 π − e c o s x cos x − e c o s x csc 2 x d x Let u = e c o s x cot x d u = − sin x e c o s x cot x − e c o s x csc 2 x d x = − cos x e c o s x − e c o s x csc 2 x d x I = [ e c o s x cot x ] 4 π 2 π = − e 2 2
