목록

125/1000, integral of x*sec^2(x), calculus 2 integration by parts

• bprp calculus b… 22일 전 2026.09.09 16:00 인기
• 6,603
  21

목록

이전글예쁜 양말에서 냄새가?그럼 영어의 문법이 보이기 시작합니다.2026.09.09
다음글American vs Japanese Math Method ???? Which one is better? ? #math #mathtricks #mathshorts2026.09.10
댓글 21

댓글목록



@mikedavis7636 　22일 전

Could the answer also be
xtan(x) + ln|cos (x)| + C?

2026-09-10 03:35

[ytcid:Ugxa2AqvpbzW_xDQ2eh4AaABAg] Could the answer also be xtan(x) + ln|cos (x)| + C?



@russelltaylor7… 　22일 전

xsec²xdx

u= x dv= sec²xdx
du = dx v=tanx

xtanx -ln|secx| + C

tanx + xsec²x -tanx = xsec²x ✔️

So xsec²xdx= xtanx-ln|secx| + C

2026-09-10 01:29

[ytcid:UgwWO54FUlydDHmcXe14AaABAg] xsec²xdx u= x dv= sec²xdx du = dx v=tanx xtanx -ln|secx| + C tanx + xsec²x -tanx = xsec²x ✔️ So xsec²xdx= xtanx-ln|secx| + C



@vikramrawat526… 　22일 전

Day 125

2026-09-09 22:52

[ytcid:UgyUnHFQF-55G1Wkze54AaABAg] Day 125



@jhouck1969 　22일 전

1/8th of the way through!

2026-09-09 22:36

[ytcid:UgxrpqruGLM0l31h1zl4AaABAg] 1/8th of the way through!



@ryanf4497 　22일 전

I plugged this into Wolfram Alpha and the result is xtan(x) + log(cos(x)) + C
Why this different?

2026-09-09 22:20

[ytcid:UgxFdmVAMXEzN6LqImN4AaABAg] I plugged this into Wolfram Alpha and the result is xtan(x) + log(cos(x)) + C Why this different?



@danmart1879 　22일 전

This guy's a math genius !!!

2026-09-09 21:42

[ytcid:UgzDosP0N7bNUldwKP14AaABAg] This guy's a math genius !!!



@BassemFanari 　22일 전

There's always another way to solve a problem ?
∫ x sec² x dx
Let t= tan x ,  dt= sec² x dx
 ⇒  x= tan⁻¹ t
∫ tan⁻¹ t dt
Do just a single IBP iteration:
Put u= tan⁻¹ t ,  dv= dt
⇒  du= 1/(t²+1) dt ,  v= t
∫ u dv = u v – ∫ v du
∫ tan⁻¹ t dt = t tan⁻¹ t – ∫ t/(t²+1) dt
∫ tan⁻¹ t dt = t tan⁻¹ t – ½ ln(t²+1) + C
Back-substitute to x:
∫ x sec² x dx = x tan x – ½ ln(tan² x + 1) + C
∫ x sec² x dx = x tan x – ½ ln(sec² x) + C
∫ x sec² x dx = x tan x – ln|sec x| + C

2026-09-09 20:38

[ytcid:UgwscJMsUaDljMcjNEF4AaABAg] There's always another way to solve a problem ? ∫ x sec² x dx Let t= tan x , dt= sec² x dx ⇒ x= tan⁻¹ t ∫ tan⁻¹ t dt Do just a single IBP iteration: Put u= tan⁻¹ t , dv= dt ⇒ du= 1/(t²+1) dt , v= t ∫ u dv = u v – ∫ v du ∫ tan⁻¹ t dt = t tan⁻¹ t – ∫ t/(t²+1) dt ∫ tan⁻¹ t dt = t tan⁻¹ t – ½ ln(t²+1) + C Back-substitute to x: ∫ x sec² x dx = x tan x – ½ ln(tan² x + 1) + C ∫ x sec² x dx = x tan x – ½ ln(sec² x) + C ∫ x sec² x dx = x tan x – ln|sec x| + C



@ImadAjjaj 　22일 전

125?

2026-09-09 19:52

[ytcid:UgzXRM5D678qBWCHCR94AaABAg] 125?



@nathanellev311… 　22일 전

My favorite answer would probably be

xtanx + ln|cosx| + C

Just because I like + over - and cos over sec

2026-09-09 16:52

[ytcid:Ugw1b6NQZ7wwg-gGlil4AaABAg] My favorite answer would probably be xtanx + ln|cosx| + C Just because I like + over - and cos over sec



@philstubblefie… 　22일 전

Or, in reduced form, 1/8.  ?

2026-09-09 16:07

[ytcid:UgyhdIOqdU-p_nsUTNl4AaABAg] Or, in reduced form, 1/8. ?