Daily Integral 16 February 2026
Yeah, I really don’t like King’s Rule. It just feels like it came out of nowhere. I was struggling until I used the hint. I guess it is very useful but I just can’t see it sometimes.
∫ 0 1 2 4 1 + 4 x 2 d x = 2 ∫ 0 1 2 1 2 ( 1 2 ) 2 + x 2 d x = 2 [ arctan ( 2 x ) ] 0 1 2 = π 2 \begin{gathered}
\int_0^\frac{1}{2} \frac{4}{1 + 4x^2}\ dx \\
= 2 \int_0^\frac{1}{2} \frac{\frac{1}{2}}{(\frac{1}{2})^2 + x^2}\ dx \\
= 2\left[\arctan(2x)\right]_0^\frac{1}{2} \\
= \boxed{\frac{\pi}{2}}
\end{gathered}
∫ 0 2 1 1 + 4 x 2 4 d x = 2 ∫ 0 2 1 ( 2 1 ) 2 + x 2 2 1 d x = 2 [ arctan ( 2 x ) ] 0 2 1 = 2 π
I = ∫ 0 π 2 sin x sin x + cos x d x = ∫ 0 π 2 sin ( π 2 − x ) sin ( π 2 − x ) + cos ( π 2 − x ) d x = ∫ 0 π 2 cos x cos x + sin x d x 2 I = ∫ 0 π 2 sin x + cos x cos x + sin x d x I = 1 2 ∫ 0 π 2 1 d x = π 4 \begin{gathered}
I = \int_0^\frac{\pi}{2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\ dx \\
= \int_0^\frac{\pi}{2} \frac{\sqrt{\sin(\frac{\pi}{2} - x})}{\sqrt{\sin(\frac{\pi}{2} - x}) + \sqrt{\cos(\frac{\pi}{2} - x})}\ dx \\
= \int_0^\frac{\pi}{2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\ dx \\
2I = \int_0^\frac{\pi}{2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\ dx \\
I = \frac{1}{2} \int_0^\frac{\pi}{2} 1\ dx \\
= \boxed{\frac{\pi}{4}}
\end{gathered}
I = ∫ 0 2 π sin x + cos x sin x d x = ∫ 0 2 π sin ( 2 π − x ) + cos ( 2 π − x ) sin ( 2 π − x ) d x = ∫ 0 2 π cos x + sin x cos x d x 2 I = ∫ 0 2 π cos x + sin x sin x + cos x d x I = 2 1 ∫ 0 2 π 1 d x = 4 π
I = ∫ 0 π 4 e x tan x ( tan x + 1 ) d x = ∫ 0 π 4 e x tan 2 x + e x tan x d x = ∫ 0 π 4 e x ( sec 2 x − 1 ) + e x tan x d x = ∫ 0 π 4 − e x + sec 2 x + e x tan x d x Let u = e x tan x , d u = e x tan x + e x sec 2 x I = − ∫ 0 π 4 e x d x + [ e x tan x ] 0 π 4 = − ( e π 4 − 1 ) + 1 = 1 \begin{gathered}
I = \int_0^\frac{\pi}{4} e^x\tan x(\tan x + 1)\ dx \\
= \int_0^\frac{\pi}{4} e^x\tan^2 x + e^x\tan x\ dx\\
= \int_0^\frac{\pi}{4} e^x(\sec^2 x - 1) + e^x\tan x\ dx\\
= \int_0^\frac{\pi}{4} -e^x+\sec^2 x + e^x\tan x\ dx \\
\text{Let $u = e^x\tan x$, $du = e^x\tan x + e^x\sec^2 x$} \\
I = -\int_0^\frac{\pi}{4} e^x\ dx + \left[e^x\tan x\right]_0^\frac{\pi}{4} \\
= -(e^\frac{\pi}{4} - 1) + 1 \\
= \boxed 1
\end{gathered}
I = ∫ 0 4 π e x tan x ( tan x + 1 ) d x = ∫ 0 4 π e x tan 2 x + e x tan x d x = ∫ 0 4 π e x ( sec 2 x − 1 ) + e x tan x d x = ∫ 0 4 π − e x + sec 2 x + e x tan x d x Let u = e x tan x , d u = e x tan x + e x sec 2 x I = − ∫ 0 4 π e x d x + [ e x tan x ] 0 4 π = − ( e 4 π − 1 ) + 1 = 1
