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Integration by Parts (Part-1)
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This lesson introduces the concept of integration by parts for jee main and advanced

Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | Best Online Maths Educator | Highest Overall Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com

U
Unacademy user
mam are these questions from 6 th August Hindu paper..
P L Bhargavi
a year ago
yes.
Yeah thnq mam. ur approach is so good mam.
Yeah thnq mam. ur approach is so good mam.
sir the problems related to substitution are very tough what should I do?
sir in 2nd example can't we take sin^-1 x as t and then solve it?
Vineet Loomba
a month ago
After that again u will hv to apply IBP ...why dont u give it a try?
Fry Pan
a month ago
yes sir I did...it came as integration of sin t. t. dt but after that answer did not match...when I applied ibp... maybe I'll give another try.... thank you for replying Sir, u have a gr8 way of teaching
Vineet Loomba
a month ago
Ans will match once u convert back t to x which is little bit difficult ..thts why i did by this method
Sir First problem mein - cos2x ki jagah +cos2x aa raha hain....Solution by me : imgur.com/a/sglzz89
Saumit Up
2 months ago
Okay sorry, at 8:10 you explained it as a Typing error
first problem in last step apne cosx ka integration nhi kiya direct answer me 2cosx type krdiya
Vineet Loomba
6 months ago
dhyan se solve kro notebook pe ... cosx ki do baar integration h ... aur bahar 2 hai ..isiliye 2 cos x aya h
User Died
6 months ago
ok ok sir i get it and thnku so much
sir in 1st example why can't we apply integration by parts on the answer that is 2xsinx -x^2cosx they both contains 2 function
Vineet Loomba
a year ago
why do u want do integration again ..answer has already come ...why to apply integration again?
Ankur Kumar
a year ago
sir because the integration has still 2 functions in it . i think that it's not completely integritared
Vineet Loomba
a year ago
Which two functions are there ..its already integrated by integration by parts ...that laat step is the answer
  1. INDEFINITE INTEGRATION INTEGRATION BY PARTS IIT-IEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)


  2. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates or search me on Google * Share among your peers as SHARING is CARING!!


  3. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Integration By Parts du dx where u & v are differentiable functions and are commonly designated as first & second function respectively. product of two functions - (first function) x (integral of the second function) Integral of [(differential coefficient of the first function) The integral of the x (integral of the second function)]" MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  4. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ILATE Rule Choose the first function as the function which comes first in the word ILATE, where; IInverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function & E-Exponential function. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  5. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration: Evaluate:x2 sin x dx Solution: x2 sin x dx = x | sin x dx-1 (2x | sin x dx)dx -x' cos x +2[x scos x dx -ascos x dx)dx =-x-cos x + 2x sin x-2 cos x + c MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  6. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) X sin 'X dx Example Find J Solution Let fint function be sin 'x and second function be a Frst we find the integral of the second f nction,ie. x dr First we find the integral of the second function, i.e. Put t=1-x". Then dt =-2xdx x dx Therefore x sin"xdx = (sin-ix)-1-X2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  7. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration: Evaluate: (cos#xdx Solution Consider I-cos Vxdx Let then- Vx dx = 2Vxdt Icost.2tdt i.e. or dx = 2t dt SO taking t as first function, then integrate it by part 1-2 | t cos tdt cos tdt dt = 2| tsin t- 1.sin tdt l = 2[tsin t-cost]-c MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  8. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration Evaluate : Jit sinx dx x(1-sinx) (1 +sinx)(1 -sin x) dxx sec xdx x sec x tan xdx Solution: Let I= dx dx 1 sin x x(1 sin x) x(1 -sinx) | 1-sin#x coS X x sec xdx - sec xdx dx x secx tan xdx- sec x tan xdx dx tan xdx-xsecx secxdx = [xtan x-enl sec x1]-[xsec x-In l sec x + tan x 1]-c x (tan x - sec x) +secx + tanx)X(1-sin x) -x 1 - sinx) = x ( tan x-sec x) + In + En 1 sin x | +c sec x coS X MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


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