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104/1000, integral of 1/(1+x^2), trig substitution

• bprp calculus b… 오래 전 2026.08.19 16:00 인기
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@vikramrawat526… 　오래 전

Day 104

2026-08-19 18:12

[ytcid:UgxiObfDI9WK-in3_Eh4AaABAg] Day 104



@ToànLê-b1h 　오래 전

Good?
∫1/(1-x²) dx=??

2026-08-19 21:11

[ytcid:UgwXH7cCcyigVI_4vHB4AaABAg] Good? ∫1/(1-x²) dx=??



@BSky2025 　오래 전

Interesting ?

2026-08-19 20:34

[ytcid:UgxFnIMStvq4eTCvKOR4AaABAg] Interesting ?



@Evan_Ebraheem 　오래 전

We should memorize this so well u cant do trigonometric integrals without it?

2026-08-19 16:23

[ytcid:UgxDFfh-YpI1XuV9Veh4AaABAg] We should memorize this so well u cant do trigonometric integrals without it?



@erynn9770 　오래 전

Now I should only know, what a sec is. Except the short form of a time unit, that is...

We only ever used sin, cos, tan in school and even university.

2026-08-20 04:35

[ytcid:UgxZqfumjp10h5UhhzZ4AaABAg] Now I should only know, what a sec is. Except the short form of a time unit, that is... We only ever used sin, cos, tan in school and even university.



@NH_RSA260_4 　오래 전

Actually, you're not integrating one.
You're integrating dθ.

2026-08-20 01:27

[ytcid:UgxgQYO5SedV6SjIBeF4AaABAg] Actually, you're not integrating one. You're integrating dθ.



@surenazand 　오래 전

Could you please post hyperbolic sector area and its integral?? It is basic but a little challenging

2026-08-19 23:47

[ytcid:UgzEjea5_4t1tX8ABFZ4AaABAg] Could you please post hyperbolic sector area and its integral?? It is basic but a little challenging



@johnnyli4993 　오래 전

Memorize this, memorize that. In an exam this is completely nonsense. I know you propose this as an exercise notwithstanding.

2026-08-19 23:07

[ytcid:Ugw_MiIzJwsar8CWnsx4AaABAg] Memorize this, memorize that. In an exam this is completely nonsense. I know you propose this as an exercise notwithstanding.



@Radhamani-t9t8… 　오래 전

Then find the derivative of arctan(x)

2026-08-19 21:49

[ytcid:UgyUdq3-gvEZUdauEz14AaABAg] Then find the derivative of arctan(x)



@YunruiHu 　오래 전

I discovered another way to do it. you can first draw a triangle with one side x and one side 1 and put your reference angle next to the side with labelled 1

now the hypotenuse is sqrt(x^2+1)

then cos(θ)= 1/sqrt(x^2+1)

that means cos^2(θ) = 1/sqrt(x^2+1) this is our integrand

then we change the differential

from our triangle we know that tan(θ)= x

that means dx = sec^2(θ)dθ

substitute all of the information into the integral

we will see that cos^2(θ) and sec^2(θ) cancel out completely leaving us integrating 1 with respect to θ

integrating that gives us θ and from one of our equations
where tan(θ) = x we find that x = tan^-1(x) + C

and we are done!

2026-08-22 03:29

[ytcid:Ugxx0IHtIJ5Yst-QsLx4AaABAg] I discovered another way to do it. you can first draw a triangle with one side x and one side 1 and put your reference angle next to the side with labelled 1 now the hypotenuse is sqrt(x^2+1) then cos(θ)= 1/sqrt(x^2+1) that means cos^2(θ) = 1/sqrt(x^2+1) this is our integrand then we change the differential from our triangle we know that tan(θ)= x that means dx = sec^2(θ)dθ substitute all of the information into the integral we will see that cos^2(θ) and sec^2(θ) cancel out completely leaving us integrating 1 with respect to θ integrating that gives us θ and from one of our equations where tan(θ) = x we find that x = tan^-1(x) + C and we are done!