# Simple Integration by substitution

John O' Meara

## Homework Statement

Find by letting $$U^2=(4 + x^2)$$the following $$\int_0^2\frac{x}{\sqrt{4 + x^2}}dx$$?
I can solve it by letting $$\mbox{x=2} tan(\theta)$$, But I want to be able to do it by substitution.

## The Attempt at a Solution

$$\frac{du}{dx}=\frac{d\sqrt{(4+x^2)}}{dx}=\frac{x}{\sqrt{4+x^2}}\mbox{, therefore du}=\frac{x}{u^\frac{1}{2}}\times dx\\$$ Therefore the integral is $$\int_{x=0}^{x=2}\frac{1}{u^\frac{1}{2}}du=$$0.26757, it should be $$2(\sqrt{2}-1)$$. Can you tell me where I went wrong. Thanks for the help.

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## Answers and Replies

Homework Helper
[$$\frac{du}{dx}=\frac{d\sqrt{(4+x^2)}}{dx}=\frac{x}{\sqrt{4+x^2}}\mbox{, therefore du}=\frac{x}{u^\frac{1}{2}}\times dx\\$$

What happened to the x in the above expression for du? Did you just drop it?

John O' Meara
There is a mistake in the equation, it should be $$\int_0^2\frac{x}{\sqrt{4+x^2}}dx$$. And the answer I then get =.5351. And thanks for the quick reply.

Homework Helper
Ah, ha. Then you want to evaluate the final integral between the u limits, not the x limits.

Homework Helper
$$\frac{du}{dx}=\frac{d\sqrt{(4+x^2)}}{dx}=\frac{x}{\sqrt{4+x^2}}\mbox{, therefore du}=\frac{x}{u^\frac{1}{2}}\times dx\\$$

And something is going awry here. A sqrt(u) would be the 4th root of 4+x^2. You are being sloppy.

John O' Meara
I know there is something awry in the calculation, maybe you can find what I did wrong. Since $$u^2=4+x^2$$ that implies $$u=\sqrt{4+x^2}$$ and $$\frac{du}{dx}$$ expression follows.

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Homework Helper
This shows du=(x*dx)/u. Not u^(1/2).

Bruk
First try substituting u=4+x²⇒du=2xdx⇒(1/2)du=xdx for the substitution part
that should be a lot easier . you are still heading the right direction
(1/2)∫(1/(u^{1/2}))du= √u
I think you can take it from here