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# More trig substitution with tangent

Another practice problem replacing x with tan(theta) in an integral. Created by Sal Khan.

## Want to join the conversation?

• It is just experience that leads to choosing an appropriate identity substitution in the beginning? It seems like picking the wrong identity will lead you down an endless corridor of trig function swapping.
• My calc book has a little table saying: "sqrt[a^2 - x^2] -> x = a sin(theta)" "sqrt[a^2 + x^2] -> x = a tan(theta)" "sqrt[x^2 - a^2] -> x = a sec(theta)" Not sure if that helps, but there's is a method of recognizing which substitution is appropriate, which boils down to determining which trig identity your integrand looks like.
• Hi at ( to about ) Sal starts by expressing that he wants to use "1+tan^2(theta) = sec^2(theta)" My question is: What do I look for in the starting integral, in order for me to find the best trig substitution.

In this example Sal chose to use "1+tan^2(theta)" but why couldn't he use "sin^2(theta)+cos^2(theta) = 1"
• Same here. Expect it also contains the useful identity sections as well, such as:
a^2-a^2 〖sin〗^2 θ= a^2 〖cos〗^2 θ)
a^2+a^2 〖tan〗^2 θ=a^2 〖sec〗^2 θ
a^2 〖sec〗^2 θ-a^2=a^2 〖tan〗^2 θ
(1 vote)
• Hi Sal,
Firstly, I have huge admiration for what you've done and what you are doing, and I think that you've begun something that will change the face of education in the world. Thank you for your huge efforts.

Now for a question :-) I'm having a spot of trouble reconciling your answer to this one with a 'standard' integral that gives 1/(a^2+u^2) = arctan(u/a)+c. Since your example is 1 / 6^2+x^2 that would imply that the integral was arctan(x/6). Am I missing something basic?

Regards,
Andrew
• Your standard integral solution is wrong. It should be integral of 1/(a^2+u^2) = (1/a)arctan(u/a)+c
• isn´t the integral just ln(36+x^2)*(1/2x)? if you differentiate that you´ll get the same integral again!
• You don't get the same integrand again. You forgot to use the product rule. The derivative of ln(36+x^2)*(1/2x) is
1/(36+x^2)-ln(x^2+36)/(2x^2)

That's because what you are doing isn't allowed. You are using the reverse chain rule when the derivative of the inside function is not present. Also you can not modify the equation to include it as you can only multiply the integrand with constants, so the x needed to use the reverse chain rule remains absent. The integrand would have to be of the form
bx/(36+x^2) (where b is a constant) in order to use the reverse chain rule.
Then the answer would be similar to what you described, (b/2)*ln(36+x^2)+C.
• The way u have done is quite complicated and involve a high risk of doing a mistake...we can just substitute "x=6tan(theta)" in the beginning and move on with it..the answer is same and saves time too
• For this, couldn't you just do 1/(36+x^2)=1/36*1/(1+1/36x^2) and then do u=1/6x and then use the inverse tan derivative to get the answer?
• Yes, you could! (And I think it's a lot easier too.) :) Good eye. I think Sal was just using this as another example of trig substitution.
(1 vote)
• can someone explain the difference b/w sin^-1=arcsin and 1/sin=csc? doesn't sin^-1=1/sin????
(1 vote)
• The first term is sin^-1(x). The second term is (sin(x))^-1
• what happens if you have a value scaling the x^2?
i.e

integral of (1/(5x^2+7))

?
• This is a little bit more complicated than shown in the video, but if we want to scale the x^2-term with integer a, you get the integral ArcTan[(Sqrt[a] x)/Sqrt[7]]/(Sqrt[7] Sqrt[a]). If we want to scale the x^2-term with 5 we get ArcTan[Sqrt[5/7] x]/Sqrt[35]. You can play around with the numbers and get integrals such as following:

a=1: ArcTan[x/Sqrt[7]]/Sqrt[7]
a=2: ArcTan[Sqrt[2/7] x]/Sqrt[14]
a=3: ArcTan[Sqrt[3/7] x]/Sqrt[21]

A pattern appear, see if you can continue with a=4 and try to understand the relations.
• Isn't it easier just to split the function? For example f(x) = 1/(36+x^2) = 1/36 + 1/x^2 = 1/36 + x^-2
Which makes F(x) = 1/36x - x^-1 + c = 1/36x - 1/x + c