Daily Integral 14 February 2026
Beginner and Easy integrals are all pretty easy. I was stuck at the medium integral due to not knowing the King’s property. This property seems quite random to me, not used to it yet.
I = ∫ 0 π 2 sin x e cos x d x Let u = cos x , d u = − sin x d x I = ∫ 0 1 e u d u = e − 1 \begin{gathered}
I = \int_0^\frac{\pi}{2} \sin x\ e^{\cos x}\ dx \\
\text{Let $u = \cos x$, $du = -\sin x\ dx$} \\
I = \int_0^1 e^u\ du \\
= \boxed{e - 1}
\end{gathered}
I = ∫ 0 2 π sin x e c o s x d x Let u = cos x , d u = − sin x d x I = ∫ 0 1 e u d u = e − 1
∫ 0 1 1 − x 2 cot ( arcsin ( x ) ) d x Simplify the denominator: cot ( arcsin ( x ) ) = cos ( arcsin ( x ) ) x = 1 − x 2 x by drawing a triangle Sub into original integrand I = ∫ 0 1 1 − x 2 1 − x 2 x = ∫ 0 1 x d x = [ 1 2 x 2 ] 0 1 = 1 2 \begin{gathered}
\int_0^1 \frac{\sqrt{1 - x^2}}{\cot(\arcsin(x))}\ dx \\
\ \\
\text{Simplify the denominator:} \\
\cot(\arcsin(x)) = \frac{\cos(\arcsin(x))}{x} \\
=\frac{\sqrt{1 - x^2}}{x} \text{ by drawing a triangle} \\
\ \\
\text{Sub into original integrand} \\
I = \int_0^1 \frac{\sqrt{1 - x^2}}{\frac{\sqrt{1 - x^2}}{x}} \\
= \int_0^1 x\ dx \\
= \left[\frac{1}{2} x^2\right]_0^1 \\
= \boxed{\frac{1}{2}}
\end{gathered}
∫ 0 1 cot ( arcsin ( x ) ) 1 − x 2 d x Simplify the denominator: cot ( arcsin ( x ) ) = x cos ( arcsin ( x ) ) = x 1 − x 2 by drawing a triangle Sub into original integrand I = ∫ 0 1 x 1 − x 2 1 − x 2 = ∫ 0 1 x d x = [ 2 1 x 2 ] 0 1 = 2 1
I = ∫ 0 π 2 x sin 2 x sin 4 x + cos 4 x d x = ∫ 0 π 2 ( π 2 − x ) sin 2 x sin 4 x + cos 4 x d x 2 I = ∫ 0 π 2 ( x + π 2 − x ) sin 2 x sin 4 x + cos 4 x I = π 4 ∫ 0 π 2 sin 2 x sin 4 x + cos 4 x = π 4 ∫ 0 π 2 sin 2 x 1 − 1 2 sin 2 2 x Let u = cos 2 x , d u = − 2 sin 2 x d x I = − π 4 ∫ 1 − 1 1 1 + u 2 d u = π 4 ∫ − 1 1 1 1 + u 2 d u = π 4 [ arctan ( u ) ] − 1 1 = π 4 ( π 4 − − π 4 ) = π 2 8 \begin{gathered}
I = \int_0^\frac{\pi}{2} \frac{x\sin 2x}{\sin^4 x + \cos^4 x}\ dx \\
= \int_0^\frac{\pi}{2} \frac{\left(\frac{\pi}{2} - x\right)\sin 2x}{\sin^4 x + \cos^4 x}\ dx \\
2I = \int_0^\frac{\pi}{2} \frac{(x+\frac{\pi}{2}-x) \sin 2x}{\sin^4 x + \cos^4 x} \\
I = \frac{\pi}{4} \int_0^\frac{\pi}{2}\frac{\sin 2x}{\sin^4 x + \cos^4 x} \\
= \frac{\pi}{4} \int_0^\frac{\pi}{2}\frac{\sin 2x}{1 - \frac{1}{2}\sin^2 2x} \\
\text{Let $u = \cos 2x$, $du = -2\sin 2x\ dx$} \\
I = -\frac{\pi}{4}\int_1^{-1}\frac{1}{1 + u^2}\ du \\
= \frac{\pi}{4}\int_{-1}^{1}\frac{1}{1 + u^2}\ du \\
= \frac{\pi}{4}\left[\arctan(u)\right]^1_{-1} \\
=\frac{\pi}{4} \left(\frac{\pi}{4} - -\frac{\pi}{4}\right) \\
=\boxed{\frac{\pi^2}{8}}
\end{gathered}
I = ∫ 0 2 π sin 4 x + cos 4 x x sin 2 x d x = ∫ 0 2 π sin 4 x + cos 4 x ( 2 π − x ) sin 2 x d x 2 I = ∫ 0 2 π sin 4 x + cos 4 x ( x + 2 π − x ) sin 2 x I = 4 π ∫ 0 2 π sin 4 x + cos 4 x sin 2 x = 4 π ∫ 0 2 π 1 − 2 1 sin 2 2 x sin 2 x Let u = cos 2 x , d u = −2 sin 2 x d x I = − 4 π ∫ 1 − 1 1 + u 2 1 d u = 4 π ∫ − 1 1 1 + u 2 1 d u = 4 π [ arctan ( u ) ] − 1 1 = 4 π ( 4 π − − 4 π ) = 8 π 2
