# Thread: integration by substitution

1. ## integration by substitution

Consider the integral of sin(2x+1).
I note the normal method would be to let u=2x+1(if you are asked to use substitution).
But can someone explain the following method:
integral (sin(2x+1)dx)=integral(sin(2x+1) d(2x+1)/2)?

2. ## Re: integration by substitution

I take that you mean this

$\int \ \ \sin(2x+1) \ \ dx = \int \ \ \sin(2x+1) \ \ \frac{d}{dx}(\frac{2x+1}{2}) \ dx$

We know that $\frac{d}{dx} (\frac{2x+1}{2})$ is 1 so they basically subbed that 1 into the integral in the right hand side.

Using the Substitution , you sub it into the equation and cancel things out with the chain rule like this

$\int \ \ \sin(u) \ \ \frac{d}{dx}(\frac{u}{2}) \ dx = \int \ \ \sin(u) \ \ \frac{d}{du}(\frac{u}{2}) \frac{du}{dx} \ dx = \\ \\ \int \ \ \sin(u) \ \ \frac{d}{du}(\frac{u}{2}) du = \\ \\ \int \ \ \sin(u) \ \ (\frac{1}{2}) du$

That's how it works, but no 3U student would get taught like that.

3. ## Re: integration by substitution

Yeah but as 'integral95' said thats a bull crap half assed answer that wont be asked to 3 unit students

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