Integration of sq. root of secx - 1 || very Important|| CBSE XII MATHS INIDA #12 special integration
Вставка
- Опубліковано 9 лют 2025
- The video Integration of sq. root of secx -1 explains the concept and fully solved the problem of special integration.
Popular And Recommended Videos:
1)How to use Log Tables Basic Knowledge part 1 of 1 :
bit.ly/2uAHvFK
2)How to use Log Tables in Solving Question Part 2 of 1 :
bit.ly/2vdpZuR
3)CBSE sample paper XII new pattern class XII fully analysed get ready to score 100 in boards 2017:
bit.ly/2hEmhp0
4)Important Question On Indefinite Integration :
bit.ly/2vGqspR
About Us:
AVTE is an Institution which was established in the year 1988. Its rapid growth is aptly attributed to it's highly professional style of coaching.
It has produced unbeatable results from the past 30years and still runs strong.
Website: www.avteschool.org
Most Popular Programs By AVTE:
TEST SERIES:
It aims to reduce the anxiety faced by a lot of students during their exams.This program is prepared under the guidance of subject experts , to give a regular practice and revision to the students.
For further inquires :
Email- avte.in@gmail.com,
avte37@yahoo.co.in
Phone- +919810934519,
01129239855
Address-bit.ly/AVTEmap
WHO EVER DID THIS VIDEO NEEDS TO DO IT OVER USING THE LESS COMPLICATED METHOD I OUTLINE ABOVE WHICH MULTIPLIES AND DIVIDES BY SQUARE ROOT OF
( SECANT (X) +1 ). THE WAY DESCRIBED HERE IS UNNECESSARILY COMPLICATED.
sir can we do it by writing secx-1 as tan^2x/secx+1 and then substituting secx=t and then substituting t+1=k^2
Which book is this
Watched and Understood
Thank You
Samarth Seth..
Sir plz solve root cosecx-1 integration
Done sir, thankyou
to integrate sqrt ( sec x - 1 ) multiply and divide by sqrt (sec x -1 ) and you have in the numerator sqrt((sec x)^2 - 1) which reduces to sqrt ((tan x)^2); then in the denominator
you will now have sqrt (sec x - 1) now multiply the numerator and denominator by sec x and you have (sec x tan x dx)/(sec x sqrt (sec x - 1). Now get rid of the trig functions
by letting u = sec x and du = sec x tan x dx. your integral now looks like du / (u(sqrt (u - 1)). Now let u - 1 = v^2 and the integral becomes
2vdv / ((v^2 + 1) sqrt(v^2)) or 2dv /(v^2 + 1) and this integral is 2 arc tan (v) + c or 2 arc tan (sqrt(u - 1) + c or 2 arc tan (sqrt(sec x - 1) + c. Note how similar the integrals of
sqrt ((sec x + 1)) which I did here is to sqrt((sec x - 1)) which I did above. This technique will also work for integrals of sqrt(csc x + 1) and sqrt(csc x - 1) This is a straightforward
approach and one that is simpler than the one shown in the video, though both are correct.
regards
buffnixx
Multiply and divide by sqrt(secx+1)*
Akash, thank you for correcting my error above.
Done sir👍🏻👍🏻
multipy and divide by sqrt(sec x + 1) then you have in numerator sqrt ((sec x)^2 - 1 ) which equals sqrt(tangent of x squared) which reduces to tan x. In the denominator
you have sqrt ((sec x + 1)) Then you multiply numerator and denominator by sec x and you get in the numerator sec x tan x dx and in the denominator sec x(sqrt (sec x + 1)
then get rid of the trig functions by substituting u = sec x and du = sec x tan x dx. your integral now looks like du/(u(sqrt(u + 1)). Now let u + 1 = v^2 and the integral now
becomes 2v dv /(v^2 - 1) sqrt(v^2) or 2dv / (v^2 - 1) and this integral is 2 arc tanh (v) + c or 2 arc tanh (sqrt(u + 1) + c or 2 arctanh (sqrt(sec x + 1) + c. A similar technique
will work for sqrt (sec x + 1) and sqrt (csc x -1 ) and sort (csc x + 1). the arc tanh function is the inverse hyperbolic tangent function. I think this approach is the most
simple and straigtforward one you can take though of course the solution shown here is also correct.
regards
buffnixx
In solving these four integrals sqrt(sec x + 1), sqrt(sec x - 1). sqrt(csc x -1), and sqrt(csc x + 1) the solution of two involve the arc tarngent function and the other
two the inverse hyperbolic tangent function. The integral of 1/(1 + x^2) is of course arc tangent x + c while 1/(1 - x^2) is either 1/2 log(1+x)/(1-x) +c or
arc tanh x + c, which is preferable as it avoids having to do partial fraction decomposition of 1/(1-x)/(1+x). The inverse hyperbolic tangent function is
sometimes called the argument inverse hyperbolic function so it is sometimes abbreviated arg tanh instead of arc tanh.
As to the solution offered here in this video, it a perfect example of choosing a complicated somewhat convoluted solution where a simple one exists. By multiplying the numerator
and denominator by sqrt(sec x +1) all you have to deal with as far as trig functions is u = sec x and du = sec x tan x dx and (tan x)^2 + 1 = (sec x)^2. The numerator the simplifies to
sec x tan x dx and the rest is easy and straightforward. Not so much so with this video! If you want to learn and remember a solution to the integral shown in this video, use my
alternative solution. Much easier to apply and remember!
Done sir, thank you!
Thank u vy much sir😃😄😃😄
Kindly do all the playlists related to XII and if you wish to do systematic classes do Whatsapp at 09871041788
Sir what is the name of that book where you pull the integrals out of?
It is a book by AVTE
interesting solution sir thank u keep it up!
35 th question 👍🏼
Nice question
sec(x)=t-tan(x)
Can we use this substitution (i didnt test it)
May be keep it up
sec(x)=t-tan(x)
and
cos(x)=(1-sin(x))t
is the same substitution
First can be used if our integrand is expressed with secx and tanx
Second can be used if our integrand is expressed with cosx and sinx
I found this substitution after i had played with Euler substitutions
In fact i found Weierstrass substitution with shifted angle
Sir can u make more important sums sir the integration playlist is awesome.thanks to u :)
Done 👍👌
Kindly also do integration of sq. root of (secx + 1) dx
to integrate sqrt ( sec x + 1 ) multiply and divide by sqrt (sec x -1 ) and you have in the numerator sqrt((sec x)^2 - 1) which reduces to sqrt ((tan x)^2); then in the denominator
you will now have sqrt (sec x - 1) now multiply the numerator and denominator by sec x and you have (sec x tan x dx)/(sec x sqrt (sec x - 1). Now get rid of the trig functions
by letting u = sec x and du = sec x tan x dx. your integral now looks like du / (u(sqrt (u - 1)). Now let u - 1 = v^2 and the integral becomes
2vdv / ((v^2 + 1) sqrt(v^2)) or 2dv /(v^2 + 1) and this integral is 2 arc tan (v) + c or 2 arc tan (sqrt(u - 1) + c or 2 arc tan (sqrt(sec x - 1) + c. Note how similar the integrals of
sqrt ((sec x + 1)) which I did here is to sqrt((sec x - 1)) which I did above. This technique will also work for integrals of sqrt(csc x + 1) and sqrt(csc x - 1) This is a straightforward
approach and one that is simpler than the one shown in the video, though both are correct.
regards
buffnixx
Superb 😊
Very helpful Indeed
Thanku sir......
Kindly watch all the playlists, and for more you may WhatsApp at 9871041788 , regards
@@avte-edu sure sir
multiply numerator and denominator by sqrt(sec x + 1)
Do rationalisation in this question by 1+cosx/1+cosx it would be more appropriate and easier
thankksss
Doen sir thank you
Excellent sir
Can we purchase this book
Yes
Aryan Philip whats the name of his book?
done sir
Thank you sir
MUCH TOO COMPLICATED!! Best to just multiple by (sqrt((sec x) + 1)/ sqrt((sec x )+ 1) and then you have (sqrt(sec x)^2 - 1)/(sqrt(sec x + 1) which is (sqrt((tan x)^2))/(sqrt((sec x) + 1)
which is tan x / (sqrt((sec x) + 1) and from this point multiply numerator and denominator by sec x and you then have ((sec x)(tan x))/((sec x)*((sqrt(sec x) + 1)). Now let sec x +1 = u^2
and sec x tan x dx = 2udu and we now have ((u du) / ((u^2 - 1)* u)) and after canceling out the u in the numerator and denominator we are left the the integral of (1 / (u*2 - 1) From this point
we can proceed with partial fractions and obtain an answer expressed in log functions or best way jut multiply inside and outside by minus one and we have
Integral = minus the integral of ( 1 - u^2) which in minus arg tanh (u) + c or minus arg tanh (sqrt(sec x + 1) + C. “arg tanh” is the inverse hyperbolic cotangent” which is sometimes
written as “arctanh” or “artanh” but by using this function you save yourself the trouble of doing partial fractions. Memorize this one and you will save yourself time if it ever pops
up on one of your tests!! (The abbreviation argtanh which is used by some authors stands for argument inverse hyperbolic tangent). The answer shown in this video is just as correct
as this one, just different. The advantage of the one I am showing you here is that it saves time. Also note that this technique can be used to solve the integrals of
sqrt (sec x + 1) and also sqrt (csc x + 1) and sqrt (csc x - 1)
solution this question integrat √secx dx
0:40 👍👍
There is an another way of solving this by taking √(secx-1)=t^2
done
THERE IS ABSOLUTELY NO EXCUSE FOR MAKING THIS INTEGRAL THIS DIFFICULT WHEN A MUCH SIMPLER AND MORE STRAIGHTFORWARD ANSWER EXISTS.
SEE BELOW.....
Put secx-1=t^2
Watching this is like trying to scratch your left ear with you right hand. See the simpler solution outlined above.....
In trying to solve an integral remember there are any number of correct solutions with different approaches. The approach here is too complicated and though correct if you are having
to solve this problem on an exam see the simpler approach I outline above.
I like your pen
my answer is different
What method did you use then, maybe i can help?
Done sir, thank you!
done sir, thanks
Done sir👍🏻
Thanks sir
Done sir
done
Done sir 👍🏻
Done sir👍🏻
Done sir ,thankyou
done sir
Done sir
Thank you
Done sir
done sir. Thanks
done sir, thankyou
Done sir
Done sir