Sunday, December 12, 2021

Six Integrals from Calculus I: Substitution vs Integration by Parts

Six Integrals from Calculus I:  Substitution vs Integration by Parts


Even slight differences integrands can change the strategy.  Observe.


Substitution


For these three integrals, the substitution u = x^2 will serve well.


Observe:

u = x^2

du = 2 x dx

du / 2 = x dx



∫ x sin x^2 dx

= ∫ 1/2 sin u du

= -1/2 cos u + C

= -1/2 cos x^2 + C


∫ x cos x^2 dx

= ∫ 1/2 cos u du

= 1/2 sin u + C

= 1/2 sin x^2 + C


∫ x e^(x^2) dx

= ∫ 1/2 e^u du

= 1/2 e^u + C

= 1/2 e^(x^2) + C


Integration by Parts


General:   ∫ u dv = u v - ∫ v du,  v and u are functions (usually of x)


∫ x^2 sin x dx

u = x^2, dv = sin x dx;   du = 2 x dx, v = - cos x


= -x^2 cos x + ∫ 2 x cos x dx

u = 2 x, dv = -cos x dx;  du = 2 dx, v = -sin x


= -x^2 cos x + 2 x sin x - ∫ 2 sin x dx

= -x^2 cos x + 2 x sin x - ∫ 2 sin x dx



∫ x^2 cos x dx

u = x^2, dv = cos x dx;   du = 2 x dx, v = sin x


= x^2 sin x - ∫ 2 x sin x dx

u = 2 x, dv = sin x dx;  du = 2 dx, v = -cos x


= x^2 sin x + 2 x cos x - ∫ 2 cos x dx

= x^2 sin x + 2 x cos x - 2 sin x + C



∫ x^2 e^x dx

u = x^2, dv = e^x dx;  du = 2 x dx, v = e^x


= x^2 e^x - ∫2 x e^x dx

u = 2 x, dv = e^x dx;  du = 2 dx, v = e^x


= x^2 e^x - 2 x e^x + ∫2 e^x dx

= x^2 e^x - 2 x e^x + 2 e^x + C

= e^x (x^2 - 2 x + 2) + C


Bottom line, being good at integration, like everywhere else in math, will require practice, experience, and maybe later in life, some retraining.  


Students, best of luck in your studies, in class and in life.  


Eddie 


All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 


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