int (ln (x+sqrt(x^2 -1))

[@chinmaykowshik5161]

the inside of the the integral at the beginning is actually inverse hyperbolic cosine x

[@miloradtomic]

It's nice.

[@AlbertTheGamer-gk7sn]

∫arccosh(x)dx
u = arccosh(x)
d(arccosh(x))/dx = 1/(sinh(arccosh(x)))
cosh^2(x)-sinh^2(x)=1
-sinh^2(x)=1-cosh^2(x)
sinh^2(x)=cosh^2(x)-1
sinh(x)=√cosh^2(x)-1
d(arccosh(x))/dx = 1/(√cosh^2(arccosh(x))-1)
= 1/(√x^2-1)
du = dx/√x^2-1
dv = dx
v = x
∫udv = uv - ∫vdu
∫arccosh(x)dx = x*arccosh(x) - ∫xdx/(√x^2-1)
= x*arccosh(x) - √x^2-1 + C

[@Red_Angel-dw4cv]

The handwriting is beautiful

[@aiersentayier6511]

Your explanation is great, I support you..😃😃

[@JourneyThroughMath]

Gonna admit, the square root spooked me so I started with a trig sub, it gave me ln(secx+tanx)

[@miloradtomic]

To be continued in all beauty...

[@raivogrunbaum4801]

Wrong substitution put x=cosh t then you get ln(sinh t+cosh t)=ln(e^t)=t

[@JourneyThroughMath]

@@raivogrunbaum4801 not terribly familiar with hyperbolic trig. Didnt think of it. Good idea though!

[@nanamacapagal8342]

​@@raivogrunbaum4801 there is no such thing as a wrong sub as long as it works

[@biswambarpanda4468]

Wonderful my great sir...long live

[@skwbusaidi]

We can set x= sect
dx = sect tant dt
Then we do integeration by part
Differciate the ln and Integerate the rest ( sect tan t)

[@holyshit922]

I found in my tables following integral
\int_{\theta}^{\pi}\frac{\sin((n+\frac{1}{2})t)}{\sqrt{2(\cos(\theta)-\cos(t))}}\mbox{d}t
Newton , you can try to calculate this integral on one of your video
I used integration by parts combined with some trigonometric identities
to get reduction formula for this integral but maybe you will find other way to calculate this integral

[@808spvrk]

Substitute x=cosθ inside the natural log becomes e^iθ and youll get -i(int θsinθ dθ) wich can easily be solved by ibp

[@jumpman8282]

At first glance this integral looked very daunting and knowing Newton I thought it was going to be another trick question!
But I couldn't come up with anything clever, so in the end I had to resort to differentiating the integrand and was surprised by how neatly it collapsed down.

[@kevinmadden1645]

Let x equal secant theta and integrate by parts. The solution is relatively straightforward .

[@Th3OneWhoWaits]

4:57 Sir, how can you cancel when the sqrt of (x^2)-1 are still being added to x?

[@PrimeNewtons]

Canceled everything. Watch again.

[@Th3OneWhoWaits]

@@PrimeNewtons Thank you. I was not watching closely enough.

[@MASHabibi-d2d]

Thanks for an other video master

[@mirzatayerejepbayev8367]

Good job

[@surendrakverma555]

Thanks Sir

[@lhopital2132]

sir please give problems based on integration of hyperbolic functions

[@Moj94]

du=dx/sqrt(x^2-1)

[@miloradtomic]

There is a little mistake. But nevermind.

[@Occ881]

I see no mistake in there

[@jumpman8282]

@@Occ881 He forgot the 𝑑𝑥 when writing 𝑑𝑢 in terms of 𝑥. Other than that, the solution was flawless.

[@JSSTyger]

​@@Occ881 He forgot to mention that he copied my test answer. Ill let it slide :D

[@qetuoa13579gjlxvn]

You are Beautiful 🎉🎉🎉

[@JosephOnyango-s6l]

Help me solve,, cos x +1 from first principle method

[@lhopital2132]

∫ cosh⁻¹x dx = x cosh⁻¹x - (√x²-1 )+c

[@AlbertTheGamer-gk7sn]

∫arccosh(x)dx
u = arccosh(x)
d(arccosh(x))/dx = 1/(sinh(arccosh(x)))
cosh^2(x)-sinh^2(x)=1
-sinh^2(x)=1-cosh^2(x)
sinh^2(x)=cosh^2(x)-1
sinh(x)=√cosh^2(x)-1
d(arccosh(x))/dx = 1/(√cosh^2(arccosh(x))-1)
= 1/(√x^2-1)
du = dx/√x^2-1
dv = dx
v = x
∫udv = uv - ∫vdu
∫arccosh(x)dx = x*arccosh(x) - ∫xdx/(√x^2-1)
= x*arccosh(x) - √x^2-1 + C

[@AlbertTheGamer-gk7sn]

∫arccosh(x)dx
u = arccosh(x)
d(arccosh(x))/dx = 1/(sinh(arccosh(x)))
cosh^2(x)-sinh^2(x)=1
-sinh^2(x)=1-cosh^2(x)
sinh^2(x)=cosh^2(x)-1
sinh(x)=√cosh^2(x)-1
d(arccosh(x))/dx = 1/(√cosh^2(arccosh(x))-1)
= 1/(√x^2-1)
du = dx/√x^2-1
dv = dx
v = x
∫udv = uv - ∫vdu
∫arccosh(x)dx = x*arccosh(x) - ∫xdx/(√x^2-1)
= x*arccosh(x) - √x^2-1 + C

[@jud.ravindra_]

Solve this int x√x²-1dx 🙏

[@lhopital2132]

∫ ln(x+ √x²-1 ) dx = ∫ cosh⁻¹x dx

[@AlbertTheGamer-gk7sn]

∫arccosh(x)dx
u = arccosh(x)
d(arccosh(x))/dx = 1/(sinh(arccosh(x)))
cosh^2(x)-sinh^2(x)=1
-sinh^2(x)=1-cosh^2(x)
sinh^2(x)=cosh^2(x)-1
sinh(x)=√cosh^2(x)-1
d(arccosh(x))/dx = 1/(√cosh^2(arccosh(x))-1)
= 1/(√x^2-1)
du = dx/√x^2-1
dv = dx
v = x
∫udv = uv - ∫vdu
∫arccosh(x)dx = x*arccosh(x) - ∫xdx/(√x^2-1)
= x*arccosh(x) - √x^2-1 + C

[@gautamchoubey1275]

Dear sir I gmail you please solve the problem 😢