MHB What is the purpose of substitution in integration?

[shamieh]

Suppose we have $$\int \frac{4}{x^2 + 4} $$

So I understand the first thing we would so is bring the constant out and do u substitution but what I don't understand is how we can make the substitution u = $$\frac{x}{2}$$ when there clearly is no $$\frac{x}{2}$$ in the problem. I also understand how to factor out a 4 in the denominator thus getting us this

$$\int \frac{1}{\frac{x^2}{4} + 1} $$

but then I don't understand how we can say u = $$\frac{x}{2}$$ then $$du = 1/2dx$$

when $$\frac{x}{2}$$ is no where in the problem..What am I missing

 

[Guest2]

... but then I don't understand how we can say u = $$\frac{x}{2}$$ then $$du = 1/2dx$$

when $$\frac{x}{2}$$ is no where in the problem..What am I missing

$\displaystyle \frac{x^2}{4} = \left(\frac{x}{2}\right)^2$

 

[MarkFL]

I would instead use the substitution:

$$x=2\tan(\theta)$$

Then you may apply a useful Pythagorean identity...

 

[shamieh]

I would instead use the substitution:

$$x=2\tan(\theta)$$

Then you may apply a useful Pythagorean identity...

I don't understand how. Can you show me?

 

[MarkFL]

I don't understand how. Can you show me?

First, compute $dx$, and then make the substitution...what do you have now?

 

[shamieh]

First, compute $dx$, and then make the substitution...what do you have now?

$$x = 2tan\theta$$
$$dx = 2sec^2\theta$$

 

[MarkFL]

$$x = 2tan\theta$$
$$dx = 2sec^2\theta$$

Correct, now substitute for $x$ and $dx$ and what do you have?

 

[shamieh]

$$\int \frac{4}{2tan^2\theta + 4} 2sec^2\theta$$

 

[MarkFL]

$$\int \frac{4}{2tan^2\theta + 4} 2sec^2\theta$$

Not quite...you want:

$$\int\frac{4}{(2\tan(\theta))^2+4}2\sec^2(\theta)\,d\theta$$

Also, I didn't notice earlier that your computation of $dx$ did not include $d\theta$.

Now, can you simplify this?

 

[shamieh]

$$\int \frac{4}{2tan^2\theta + 4} 2sec^2\theta$$

$$\frac{8}{2} \int \frac{sec^2\theta}{tan^2\theta + 4} \, d\theta$$
$$
u = tan^2\theta + 4$$
$$du = sec^2\theta \, d\theta$$ ?

 

[Guest2]

It seems you only squared the tan and not the 2 before it as well.

 

[MarkFL]

$$\frac{8}{2} \int \frac{sec^2\theta}{tan^2\theta + 4} \, d\theta$$
$$
u = tan^2\theta + 4$$
$$du = sec^2\theta \, d\theta$$ ?

This is what I suggest...first we have:

$$2\int\frac{4\sec^2(\theta)}{4\tan^2(\theta)+4}\, d\theta$$

Now, divide each term in the numerator and denominator of the integrand by 4:

$$2\int\frac{\sec^2(\theta)}{\tan^2(\theta)+1}\, d\theta$$

Now, can you rewrite the denominator of the integrand using a Pythagorean identity?

 

[shamieh]

When you say re write it as a pythagorean identity you mean do a substitution right?

 

[MarkFL]

When you say re write it as a pythagorean identity you mean do a substitution right?

Well, it is a substitution of sorts, but not like a typical $u$-substitution. Begin with the well-known Pythagorean identity:

$$\sin^2(\theta)+\cos^2(\theta)=1$$

and divide through by $\cos^2(\theta)$...what do you get?

 

[shamieh]

yea but why do i need to do sin + cos when you can make the U substitution for tan and sec?

 

[Guest2]

yea but why do i need to do sin + cos when you can make the U substitution for tan and sec?

That will take you back to square one. You would exactly be reversing what you did.

Do you know what $\tan^2{x}+1$ is equal to? If no, then perhaps you didn't give MarkFL's question much thought; if yes, then where you have $\tan^2{x}+1$ in the integral replace that with what it equals and there isn't much left.