# How do you know when to use u-substitution for integration and what u is? (example included) (1 Viewer)

#### catha230

##### New Member
Hi, the title pretty much speaks for itself. There are so many methods to use in 4u integration so how do you identify when to use u-substitution and what the value of u is? I came across the following question (part c) in which it already hints at using u-substitution and gives you the suitable value of u. However I wonder if the question did l not, how would you manage to solve it? Especially in this case, the given value of u is 1/x which is a bit unexpected to me.
Thank you so much for taking the time reading this!

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#### Drdusk

##### π
Moderator
Hi, the title pretty much speaks for itself. There are so many methods to use in 4u integration so how do you identify when to use u-substitution and what the value of u is? I came across the following question (part c) in which it already hints at using u-substitution and gives you the suitable value of u. However I wonder if the question did l not, how would you manage to solve it? Especially in this case, the given value of u is 1/x which is a bit unexpected to me.
Thank you so much for taking the time reading this!
What I used to do in 4u is always start by thinking of a u sub and then trig sub, partial fractions and then by parts etc. I would do this in order and usually what happens is hard integrals require a sophisticated u sub or trig sub so if at first it isn't obvious what method to go with try think of a u sub or trig sub.

What usually happens is after you've practiced a lot you tend to develop an intuition for certain integrals so you will most likely on first or second try know which kind of method to choose.

I'll leave the Math gods of Bos to actually go into detail on your question.

#### KingOfActing

##### lukewarm mess
Practice makes perfect as usual! There isn't a "right" or "wrong" u-substitution. The goal of u-substitution is to simplify the integral into terms that you are more familiar with.

We can think about what substitutions would make/would not make our lives easier - for example, if we tried to simplify out the inside of the square root with $\bg_white u = x^2 - 1$, we would have $\bg_white du = 2x dx$ which is not super useful to us (we would have to write $\bg_white x$ in terms of $\bg_white u$, and then we just get square roots again).

My next instinct would then be to try simplify the inside using trigonometry. Since we have $\bg_white x^2 - 1$, I would substitute $\bg_white x = \sec \theta$. This simplifies the integral into simply $\bg_white \int \,d\theta = \theta = \sec^{-1} x$.

This is just as "correct" as using the substitution they have used - substitution doesn't actually do anything to the integral (assuming some conditions on the domain of integration) it just rewrites the integral in a form that is (hopefully) easier for us to work with.

#### catha230

##### New Member
What I used to do in 4u is always start by thinking of a u sub and then trig sub, partial fractions and then by parts etc. I would do this in order and usually what happens is hard integrals require a sophisticated u sub or trig sub so if at first it isn't obvious what method to go with try think of a u sub or trig sub.

What usually happens is after you've practiced a lot you tend to develop an intuition for certain integrals so you will most likely on first or second try know which kind of method to choose.

I'll leave the Math gods of Bos to actually go into detail on your question.
Practice makes perfect as usual! There isn't a "right" or "wrong" u-substitution. The goal of u-substitution is to simplify the integral into terms that you are more familiar with.

We can think about what substitutions would make/would not make our lives easier - for example, if we tried to simplify out the inside of the square root with $\bg_white u = x^2 - 1$, we would have $\bg_white du = 2x dx$ which is not super useful to us (we would have to write $\bg_white x$ in terms of $\bg_white u$, and then we just get square roots again).

My next instinct would then be to try simplify the inside using trigonometry. Since we have $\bg_white x^2 - 1$, I would substitute $\bg_white x = \sec \theta$. This simplifies the integral into simply $\bg_white \int \,d\theta = \theta = \sec^{-1} x$.

This is just as "correct" as using the substitution they have used - substitution doesn't actually do anything to the integral (assuming some conditions on the domain of integration) it just rewrites the integral in a form that is (hopefully) easier for us to work with.
Thank you for the insight!

#### stupid_girl

##### Active Member
I also find the given substitution unexpected. I would straight away think of the trig substitution x=sec t.

#### catha230

##### New Member
I also find the given substitution unexpected. I would straight away think of the trig substitution x=sec t.
Thank you for your reply. Could you please share with us some of your tips for integration? You seem to be an expert in this area.

#### stupid_girl

##### Active Member
Thank you for your reply. Could you please share with us some of your tips for integration? You seem to be an expert in this area.
As mentioned by DrDusk, with a lot of practice, you will develop an intuition for certain integrals.

For u-substitution, quite often you can factor out the derivative of a familiar function.
$\bg_white \int\frac{\cos x\ dx}{\sqrt{\sin x}}=\int\frac{d\left(\sin x\right)}{\sqrt{\sin x}}$
$\bg_white \int\frac{2x dx}{\sqrt{x^{2}+1}}=\int\frac{d\left(x^{2}+1\right)}{\sqrt{x^{2}+1}}$
$\bg_white \int\frac{3x^{2} dx}{\sqrt{1-x^{6}}}=\int\frac{d\left(x^{3}\right)}{\sqrt{1-\left(x^{3}\right)^{2}}}$

Also, when there is a composite function, you may try to take the inner function as the substitution.
$\bg_white \int\sin\sqrt{x}dx=\int2u\sin u\ du\ \left(u=\sqrt{x}\right)$
$\bg_white \int\sin\left(\ln x\right)dx=\int e^{u}\sin u\ du\ \left(u=\ln x\right)$

Of course, there are harder substitutions that are not obvious. I doubt you will get these in HSC.
$\bg_white \int\sqrt{\tan x}dx=\int\frac{2u^{2}}{u^{4}+1}du\ \left(u=\sqrt{\tan x}\right)$
$\bg_white \int\sqrt[3]{\tan x}dx=\int\frac{3u^{3}}{u^{6}+1}du\ \left(u=\sqrt[3]{\tan x}\right)$

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#### mathsbrain

##### Member
As mentioned by DrDusk, with a lot of practice, you will develop an intuition for certain integrals.

For u-substitution, quite often you can factor out the derivative of a familiar function.
$\bg_white \int\frac{\cos x\ dx}{\sqrt{\sin x}}=\int\frac{d\left(\sin x\right)}{\sqrt{\sin x}}$
$\bg_white \int\frac{2x}{\sqrt{x^{2}+1}}=\int\frac{d\left(x^{2}+1\right)}{\sqrt{x^{2}+1}}$
$\bg_white \int\frac{3x^{2}}{\sqrt{1-x^{6}}}=\int\frac{d\left(x^{3}\right)}{\sqrt{1-\left(x^{3}\right)^{2}}}$

Also, when there is a composite function, you may try to take the inner function as the substitution.
$\bg_white \int\sin\sqrt{x}dx=\int2u\sin u\ du\ \left(u=\sqrt{x}\right)$
$\bg_white \int\sin\left(\ln x\right)dx=\int e^{u}\sin u\ du\ \left(u=\ln x\right)$

Of course, there are harder substitutions that are not obvious. I doubt you will get these in HSC.
$\bg_white \int\sqrt{\tan x}dx=\int\frac{2u^{2}}{u^{4}+1}du\ \left(u=\sqrt{\tan x}\right)$
$\bg_white \int\sqrt[3]{\tan x}dx=\int\frac{3u^{3}}{u^{6}+1}du\ \left(u=\sqrt[3]{\tan x}\right)$
Can you please explain how this following notation make sense?
$\bg_white \int\frac{\cos x\ dx}{\sqrt{\sin x}}=\int\frac{d\left(\sin x\right)}{\sqrt{\sin x}}$
Usually we have dx at the end which means with respect to x, but what does with respect to sinx mean?

#### InteGrand

##### Well-Known Member
Can you please explain how this following notation make sense?
$\bg_white \int\frac{\cos x\ dx}{\sqrt{\sin x}}=\int\frac{d\left(\sin x\right)}{\sqrt{\sin x}}$
Usually we have dx at the end which means with respect to x, but what does with respect to sinx mean?
$\bg_white \noindent \int f(\sin x)\, d(\sin x) essentially just means \int f(u)\, du where u=\sin x. If you don't like using d(\sin x), just write it using a u-substitution, integrate with respect to u, and then write the answer in terms of x.$

$\bg_white \noindent For example, to find \int e^{\sin x}\, d(\sin x), let u=\sin x, then \int e^{\sin x}\, d(\sin x)= \int e^u\, du = e^u + C = e^{\sin x}+C.$

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