Putting The Chain Products Together


Johnny could solve simple integration problems

Johnny was happy that now he could solve simple integration problems.  In fact, he was excited and found several problems online and attempted them.  Some of them were straightforward and he whizzed through them.  There were others he got stuck with.  He mentioned it to Sara because he knew that she was always two steps ahead of him.

Johnny:  I did lots of online problems on integration.  I was happy that I could do them but some of them were tough and I did not know how to approach them.

Sara: The trick is to use more of what you learned about derivatives.  We can talk about two approaches.  First, do you remember the chain rule in differentiation?

Johnny:  Remember Sara, I got an A plus in Calculus I.  Of course, I remember the chain rule.  Here it is:

dy/dx = (dy/du). du/dx

Substitutions based on chain rule

Sara:  Hey, that’s good.  Just like chain rule for derivatives, you can also use a similar substitution in integration.  This is interesting and sometimes challenging because your skills lie in finding a substitution. The goal is to convert a more complicated ∫y/dx to a simpler ∫udu. Here is an example:

Because it could be complicated to solve ∫3x2 (x3+5)9 dx, we can use the integration by substitution. We can do the substitution u = x3+5 because du/dx = 3x2 or du = 3x2 dx.

Then  ∫3x2 (x3+5)9 dx = ∫ (x3+5)9 3x2dx = ∫ u9 du = u10/10 + c = (x3+5)10/10 + c.

Johnny:  I got it.  Give me one to do now.

Sara: Okay, this one may be harder.  What is ∫(tan x . sec x)dx?

Johnny thought for a few minutes and then said: let me try rewriting u = cos x.

Then du/dx = -sin x or du  = – sin x.dx. Therefore,

∫(tan x .sec x)dx = ∫ (sin  x)/(cos2 x)dx =∫ – 1/u2 du = 1/u + C =  1/cos x + c = sec x + c

Sara checked the answer for any errors. There were none.  She gave Johnny a big hug because if the big lug could think of this complicated solution, he would certainly head for an A plus in Calculus 2.

Johnny:  I thought you were going to tell two types of methods,  What is the second one?

Substitutions based on product rule

Sara:  Oh yes, the  second one.  Do you remember the product rule for differentiation.

Johnny:  Yes that was d(uv)/dx = udv/dx + vdu/dx

Sara:  Then you can also write ∫d(uv)/dx = ∫udv/dx + ∫vdu/dx or uv = ∫udv/dx + ∫vdu/dx or uv = ∫udv/dx + ∫vdu/dx  ot ∫udv/dx = uv- ∫vdu/dx

Johnny:  Yes, you can but show me how to use it for integration.

Sara:  Let us do this integral


Let us substitute u = x and dv = cos x dx  so that we have to integrate udv

Then du/dx =1, and v = sin x, uv = x sinx, and ∫vdu = cos x

∫udv  = uv- ∫vdu   = x sin x + cos x + c where c is a constant.

Johnny: We should leave the rest for tomorrow because I have to go home now.

Johnny tried a few more practice questions from the Calculus 2 book.  He thought that it was challenging but he liked it. The next day, as usual, Sara and Johnny got together worked on what Johnny had found more challenging.


Solve for ∫x2sin(10x)dx

Solution: Let us substitute u = x2   and  dv = sin (10x)dx

Then du/dx or du = 2x and v =  – cos (10x) / 10 x

∫x2sin(10x)dx = ∫uvdx  and ∫udv/dx = uv- ∫vdu/dx  =

– x2cos (10x) / (10 x) +((1/5)(xsin(10x) /10)-(1/10) ∫sin(10x)dx) + c =

– x2cos (10x) / 10 x +((1/5)(xsin(10x) /10)+1/100 cos(10x)) + c =

– x2cos (10x) / 10 x  + (xsin(10x)) /50+ (cos(10x))/500+ c

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