I am still enjoying your videos though I am retired professional (E & P business) since you're fully conversant with the subject matter and also capable of explaining the same so nicely.Because of your such teaching skills (articulation) you have been able to make each and every topic / problem of discussion very much interesting and easily comprehensible.Thank you so much and salute to you Sir!
I'm not a pro in math but i loved integration by parts back when i was in high school. I didn't remember how to use trigonometric substitution until now. Thank you very much! Subscribed!
you could have just used integration by parts in the first place in stead of the u-substitution. there is a x⁰ there. Using LIATE we will see that the simplifies very smoothly.
Myself I solved by sub t=arcsec(sqrt(x)) Sec t = sqrt(x) sec^2 t = x 2 sec^2 t tant = dx We can use integration by part for 2t sec^2 t tant D is 2t and I is sec^2t tan t And we get x sec(sqrt(t)) - sqrt(sqrt(t)) + c
That's a really horrible notation. The secant is the reciprocal of cosine so writing sec^-1 could equally be interpreted as cosine, i.e the multiplicative inverse NOT the functional inverse. There is no way to tell which is being used except whatever is customary in your mathematical microcosm.
I personally prefer the arc notation but using ^(-1) isn't really an issue because everyone will understand what you mean anyway. No one is gonna write sec^(-1)(x) if they mean the reciprocal of sec(x); they will just write cos(x). sec^(-1)(x) is for the most part understood to be the inverse function of sec(x)
If it’s too hard for you to distinguish between two completely different positions of a ^-1, you probably shouldn’t be concerning yourself with trig functions.
I rewrote sec⁻¹(√𝑥) as cos⁻¹(1 ∕ √𝑥) and 𝑢-substituted the entire thing. This gave me ∫𝑢⋅2 sin(𝑢) ∕ cos³(𝑢)⋅𝑑𝑥 Then I did integration-by-parts by integrating 2 sin(𝑢) ∕ cos³(𝑢) twice, which was fairly straightforward. In the end I arrived at 𝑥 cos⁻¹(1 ∕ √𝑥) − √(𝑥) sin(cos⁻¹(1 ∕ √𝑥)) - - - Because cos⁻¹(1 ∕ √𝑥) is strictly positive, we can write sin(cos⁻¹(1 ∕ √𝑥)) = √(1 − cos²(cos⁻¹(1 ∕ √𝑥)) = √(1 − 1 ∕ 𝑥), 𝑥 > 0. Thus, the answer can also be written as 𝑥 cos⁻¹(1 ∕ √𝑥) − √𝑥⋅√(1 − 1 ∕ 𝑥), which simplifies to 𝑥 cos⁻¹(1 ∕ √𝑥) − √(𝑥 − 1) Finally, we remember that cos⁻¹(1 ∕ √𝑥) = sec⁻¹(√𝑥), which gives us 𝑥 sec⁻¹(√𝑥) − √(𝑥 − 1)
Love your videos sir!! I am from 12th class and it is very helpful to learn these integration tricks!! Love from India🇮🇳
I am still enjoying your videos though I am retired professional (E & P business) since you're fully conversant with the subject matter and also capable of explaining the same so nicely.Because of your such teaching skills (articulation) you have been able to make each and every topic / problem of discussion very much interesting and easily comprehensible.Thank you so much and salute to you Sir!
I'm not a pro in math but i loved integration by parts back when i was in high school. I didn't remember how to use trigonometric substitution until now. Thank you very much! Subscribed!
شكرا لك شرح ممتاز جداً الى درجة جعلت المسائل تنشرب
Actually, when I do these problems I don't really look at them this way. Thank you Sir for making it this simple
You should do a prevention hotline promotion.
Hey kids, never stop living. Those who stop living, stop learning.😃
When doing u substitution, it is easier sometime to isolate x
u= sqrt(x)
u^2=x
2u du =dx
God! This is scarying me bro, tysm cause ur teaching me a lot
Beautiful voice sir! 😂
9:45 you can always write this expression as 1/2.(d(u^2-1))/√(u^2-1) instead of substitution. Like int(dy/√y)=2√y...it saves time.
you could have just used integration by parts in the first place in stead of the u-substitution. there is a x⁰ there. Using LIATE we will see that the simplifies very smoothly.
I love this ❤️
Myself I solved by sub t=arcsec(sqrt(x))
Sec t = sqrt(x)
sec^2 t = x
2 sec^2 t tant = dx
We can use integration by part for 2t sec^2 t tant
D is 2t and I is sec^2t tan t
And we get
x sec(sqrt(t)) - sqrt(sqrt(t)) + c
12:49 He sings. Perhaps he feels happy.
I did this by just doing integration by parts from the start and got
x*arcsec(sqrt(x)) - ½∫dx/sqrt(x-1)
=x*arcsec(sqrt(x)) - sqrt(x-1) + c
🎉
Sir please make a full basic to advance free calculas course😢
That's a really horrible notation.
The secant is the reciprocal of cosine so writing sec^-1 could equally be interpreted as cosine, i.e the multiplicative inverse NOT the functional inverse. There is no way to tell which is being used except whatever is customary in your mathematical microcosm.
I personally prefer the arc notation but using ^(-1) isn't really an issue because everyone will understand what you mean anyway. No one is gonna write sec^(-1)(x) if they mean the reciprocal of sec(x); they will just write cos(x). sec^(-1)(x) is for the most part understood to be the inverse function of sec(x)
If it’s too hard for you to distinguish between two completely different positions of a ^-1, you probably shouldn’t be concerning yourself with trig functions.
I rewrote sec⁻¹(√𝑥) as cos⁻¹(1 ∕ √𝑥) and 𝑢-substituted the entire thing.
This gave me ∫𝑢⋅2 sin(𝑢) ∕ cos³(𝑢)⋅𝑑𝑥
Then I did integration-by-parts by integrating 2 sin(𝑢) ∕ cos³(𝑢) twice, which was fairly straightforward.
In the end I arrived at
𝑥 cos⁻¹(1 ∕ √𝑥) − √(𝑥) sin(cos⁻¹(1 ∕ √𝑥))
- - -
Because cos⁻¹(1 ∕ √𝑥) is strictly positive, we can write
sin(cos⁻¹(1 ∕ √𝑥)) = √(1 − cos²(cos⁻¹(1 ∕ √𝑥)) = √(1 − 1 ∕ 𝑥), 𝑥 > 0.
Thus, the answer can also be written as 𝑥 cos⁻¹(1 ∕ √𝑥) − √𝑥⋅√(1 − 1 ∕ 𝑥),
which simplifies to 𝑥 cos⁻¹(1 ∕ √𝑥) − √(𝑥 − 1)
Finally, we remember that cos⁻¹(1 ∕ √𝑥) = sec⁻¹(√𝑥),
which gives us 𝑥 sec⁻¹(√𝑥) − √(𝑥 − 1)