One Powerful Integration Weapon

In this video, I am evaluating a very nice viewer suggested integral using interesting trigonometric substitution, instead of using u-substitution. This is a very interesting integral to be nicely evaluated using trigonometric substitution.
#math #maths
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[@gunhasirac]

nice solution but I'm triggered by the arrow notation. Tempting to take a point off lol

[@drpkmath12345]

Oops haha funny! Thanks for the support my friend👍👍👍

[@doctorb9264]

Excellent Integration problem.

[@drpkmath12345]

Thank you so much for the support my friend👍👍👍

[@domedebali632]

You are always publishing the best videos

[@drpkmath12345]

Thanks a lot my friend for your support haha👍👍👍

[@LITHICKROSHANMS-gw2lx]

Super solution sir
Sir can you make a seperate video for advanced integral techniques and upload even more comprehensive integral sir !?
Thanking you sir!!

[@drpkmath12345]

Sounds like an idea my friend! Thanks for your support👍👍👍

[@MrGLA-zs8xt]

Perfect video professor

[@drpkmath12345]

Thanks a lot my friend for your support 👍👍👍

[@mathnerd5647]

Another great video

[@drpkmath12345]

Thanks a lot my friend for your support 👍👍👍

[@k_wl]

if im not wrong this was asked in jee, trig substitution can be used in just about any function with square roots in them

[@mathnerd5647]

I vaguely remember this in jee advanced, but used u-subs instead, which also worked

[@k_wl]

@@mathnerd5647 yea i just know its in jee cuz it is usually a class illustration

[@drpkmath12345]

Haha I didnt know that👍👍👍

[@iqtrainer]

Dang this is a very interesting video professor🎉

[@drpkmath12345]

Thanks for your support my friend👍👍👍

[@Min-cv7nt]

so smart and good looking

[@drpkmath12345]

Haha thanks a lot my friend👍👍👍

[@raghvendrasingh1289]

❤
second method
put x = 3t , dx = 3dt new interval [0,1]
we have to integrate 3 { t^(1/2) } {(1-t)^(-1/2) }
or 3 { t^(3/2-1) } { t^(1/2-1) }
I = 3 gamma(3/2) gamma (1/2)/ gamma (2)
= 3(1/2) √π √π/ (1!)
= 3π/2

[@MrGLA-zs8xt]

like a quarter circle?

[@drpkmath12345]

Thats very nice my friend! Haha thanks for the comment👍👍👍

[@raghvendrasingh1289]

@@drpkmath12345 🙏

[@raghvendrasingh1289]

@@MrGLA-zs8xt yes we can evaluate it by quarter circle also
let 3 - x = t^2
d x = - 2 t dt
now integral is
2 sqrt (3 - t^2) in [0,√3]
= 2 area of quarter circle with radius √3
= 2 π(3)/4 = 3 π/2

[@tarentinobg]

You are making this too complicated
Just go straight to the x = 3 Sin^2 t. where t = theta. You don't need the preamble with the a and b variables.
Then factor out the √3 to get 1 - Sin squared and so the denominator becomes √3 cox t.
That is so much easier than what you are doing man.

[@MrGLA-zs8xt]

Why? Why x is 3sin^2t? Why? You need to explain it. That's what Dr PK did. You are skipping so many steps and say your method is easier? Is that even math? DR PK method is far more superior than yours man. Stop doing it

[@iqtrainer]

Yours is a lot uglier and off the track. You didnt even get the answer. You did it worse than dr pk

[@ginonapoli7929]

@@MrGLA-zs8xt Really?
I did the same thing. I just skipped the √a^2-b^2 part.
You asked me to explain. Okay.
If you get "1 - Sin^2 t" you can use the trig identity to get Cos^2 t.
Since you can factor out 3, you substitute x = 3 Sin^2 t
Notice this explanation just uses a high school trig pythagorean identity rather than fitting into a pattern with variables a and b

[@MrGLA-zs8xt]

@@ginonapoli7929 I did the same thing? You are that tarentinobg. Why are you pretenting like a different person? Also, I hate trig. So, your method is worse. So, Dr. PK's method is a lot more sophisticated. Easier for you. But worse for me. Plus, you keep saying trig identity I know. But why is it easier to use for that integral? You did not explain any rationale.