Important Question On indefinite Integration | classXII | cbse exams

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The given question belongs to substitution and is important for 4 mark questions in CBSE, and also important for other equivalent examination; for watching the clip one can be more motivated and confident to do questions on Indefinite Integration
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I am lucky that You uploaded such a awsm

withspobby
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Putting x=(tant)^4 will also do it easily, as it is one of the standard trick which most of the times work

ritikaransubhe
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Euler substitutions give nice result
Firstly use third Euler substitution (with the roots)
sqrt(x-x^2)=xt
Secondly use first Euler substitution (with leading coefficient)
sqrt(1+t^2)=u-t

holyshit
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Sir you are my best teacher in integral.

technicallightingfriend
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Yes but i had another way to solve this integral, here are the steps:
I = int of dx/(1+sqrt(x))*sqrt(x-x^2), now factor out x from sqrt(x-x^2):
I = int of dx/(1+sqrt(x))*sqrt(x(1-x))
I = int of
Let u=sqrt(x)
du=dx/2*sqrt(x) so the integral becomes:
I = 2*int of du/(1+u)*sqrt(1-u^2)
Let u=sin(t)
du=cos(t)*dt so the integral becomes:
I = 2*int of
I = 2*int of
I = 2*int of cos(t)*dt/(1+sin(t))*cos(t), the 2 cosines cancel:
I = 2*int of dt/(1+sin(t)), multiply and divide by 1-sin(t):
I = 2*int of (1-sin(t))*dt/(1-sin^2(t))
I = 2*int of (1-sin(t))*dt/cos^2(t)
I = 2*int of sec^2(t)*dt - 2*int of tan(t)*sec(t)*dt
I = 2*[tan(t)-sec(t)] = 2*[u/sqrt(1-u^2)-1/sqrt(1-u^2)], so the final solution is:
I = 2*(sqrt(x)-1)/sqrt(1-x) + C.

ernestschoenmakers
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sir I thought of another method I.e -> take x^1/2 common. Apply a^2-b^2 in deno. put x^1/2 = T and then take T ^2 common in deno and put 1/T^2 = y and the ans is really simple to get.

ayushhearthacker
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mukesh sir please upload all such problems and help us to ace the cbse board examination...

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THE BEST WAY TO SOLVE THIS FOR SURE )THANKS SIR)

dhruvgoel
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Its a very good question...thank you sir.
Make my mind a bit active.

sagarmandal
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Superb video!
Studied abroad I suppose your accent is cool!

Sndbest