## Vineet Loomba is teaching live on Unacademy Plus

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)

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!!

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)

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)

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)

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)

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)

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)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ENROL for this Course (Free) Enroll 439 Recommend Lessons (Like) E . Rate and Review the Course .Comments . Sharing with friends 15 68 26 ratings 7 reviews Share MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)