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Integration by Parts (Part-3)
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This lesson concludes the discussion on integration by parts with more difficult and. Different set of problems

Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE | 6 Million Minutes Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com

U
Unacademy user
sir apne inverse wala question mai 2x + 2 = 3 tan(theta) pakra kiu ? o wala samaj nehi aya plzz thora bata dijiye
UD
sir NDA ke liye bhi video banao
Vineet Loomba
2 months ago
All these can be used for nda
AD
sir meine e questions samaj li Hein....sin or sin inverse 1 ho jaye gi...tnk u so much sir🥰🥰🥰😍😍😍
Vineet Loomba
5 months ago
Welcm
AD
Aparajita Das
5 months ago
sir AK questions tha
Vineet Loomba
5 months ago
Yes ask
AD
Aparajita Das
5 months ago
sir integration ke liya or video banaye to achha hoga...🥰❣️
AD
sir eha per to sin inverse ki ans hi likha nahi gayi..or Apne to sirf dx ka ans kiya Hein..or is ans mujhe confused lag Raha Hein..e 1st bali ? Hein..to app plz mujhe samjha di ji aa..plz sir..mujhe e ? solve karna hi Hein..e ans galat Hein...yeh phir mein galat hu...plz ans me sir...e bohot accha ? Hein...so plz ans me sir..🥰🥰
Lk
dont on side sir i m not able to understand
Vineet Loomba
5 months ago
Wht is ur difficulty ..tell me i will help u :)
Lk
Lav kumar
5 months ago
sir i cant understand on slide you should make step by step on white board or paper
Vineet Loomba
5 months ago
Some courses are on slide some on whiteboard ..all new videos I am writing side by side like live writing :)
  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) 2x +2 Illustration: Evaluate: sin 4x- +8x +13 2x +2 1= sn'l- Solution: 4x2 +8x +13 2x +2 2x+2 -1 Here sin1 s n 4x2 +8x +13 2x +2) +9 Put 2X + 2 = 3 tan 2x +2 (2x + 2)2+9 3 tan Also 3sec + C MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  6. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Find [e" sin x dx Solution Take e as the first function and sin x as second function. Then, integrating by parts, we have 1-Ier sin x dx=e"(-cos x) +le'cos x dr e cos x + I, (say) Taking e and cos x as the first and second functions, respectively, in I, we get I, = ex sin x-1 e'sin x dx I=-er cos x + er sin x-1 or 21 = er (sin x-cos x) -Iex sin x dr = e(sin x-cos x) + C Substituting the value of I in (1), we get , Hence MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  7. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 21dr Example Find | | |og (log x) + (logx) Solution Let Ilog (logx) + (logx) -log (log x) dx+ (logx) In the first integral, let us take 1 as the second function. Then integrating it by parts, we get dx 1 = x log (log x)- x dx+ x log x (log x)2 dx xlog (logx)+ MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  8. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Again, consider , take 1 as the second function and integrate it by parts, we haveJ logx log* (log x) r Putting (2) in (1), we get dx dx 1- xlog (logx)-logx(logx = xlog (log x) + C x)2 J (log x) (log.r) log x MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


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