Must be a definite integral from 0 to 1. That way, you can substitute x = tan p and subsequently get the same form as in the video. At High school level, I dont think there is a way to solve this indefinite integral.
@@ernestschoenmakers8181 Joint Entrance Examination That's the main examination in India for entering Engineering colleges in the country. It's taken by students who have just cleared grade 12 in school. I should've known, you are not an Indian, right?
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I did it in a different way. I used the fact that tan=sin/cos and got ln((sin(x)+cos(x))/cos(x)) ==> ln(sinx+cosx)-ln(cos(x)) using harmonic addition, I was left with ln(sqrt(2)cos(x-pi/4))-ln(cosx) ==>ln(sqrt(2))+ln(cos(x-pi/4))-ln(cos(x)) using the substitution pi/4- x and the fact that cos(-x)=cos(x), the last 2 terms ancle out and I'm left with int from 0 to pi/4 of ln(sqrt(2)), which is trivial
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3:07 Incorrect. When no base is given by default it is base 10, not e, otherwise it would be ln(something) instead of log(something)
Throughout class 12 whatever the log is it’s supposed to be base e that’s what the book says (NCERT) he has mentioned “class 12” in his caption
In cbse log without base is base e whereas ln implies base e , regards
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Thank you Sir, your very easy method solve to question
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And makes the video for differential equation of first order
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convergent,divergent,ocillate , conditionally Convergent way
Integration 0 to Pi/4 (1 + sinX ) dX / cos^2 X
Simply write cos sq x as 1 - sin ^2 x and proceed
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Samarth Seth..
Solve this indefinite integral: ln(1+x)/(1+x^2).
Must be a definite integral from 0 to 1. That way, you can substitute x = tan p and subsequently get the same form as in the video. At High school level, I dont think there is a way to solve this indefinite integral.
Are you preparing for JEE?
@@simba2798 What is JEE?
@@ernestschoenmakers8181 Joint Entrance Examination
That's the main examination in India for entering Engineering colleges in the country. It's taken by students who have just cleared grade 12 in school.
I should've known, you are not an Indian, right?
@@simba2798 No i'm a Dutchman from the Netherlands.
Done sir thankyou
Done sir
Thank you
Thanks...
Nice question sir! Thankyou :))
Thank you very much sir
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Nice sir
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Thank u
Done sir!
a lot of thanks
done sir
Very helpfull
Done sir👍🏻👍🏻
Done goood question
I did it in a different way. I used the fact that tan=sin/cos and got ln((sin(x)+cos(x))/cos(x)) ==> ln(sinx+cosx)-ln(cos(x))
using harmonic addition, I was left with
ln(sqrt(2)cos(x-pi/4))-ln(cosx)
==>ln(sqrt(2))+ln(cos(x-pi/4))-ln(cos(x))
using the substitution pi/4- x and the fact that cos(-x)=cos(x), the last 2 terms ancle out and I'm left with
int from 0 to pi/4 of ln(sqrt(2)), which is trivial
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Done sir :)
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No doubt you are an expert on this topic, however, unfortunately there is a lack of explaining things as you go.
Revised sir
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