Integrating by Substitution

This (hopefully) shouldn't take too long to explain. Basically, you generally use integrating by substitution when you've got something in brackets raised to a power and something else sitting outside those brackets that isn't a multiple of the derivative of the stuff in the brackets.

When you integrate by substitution, you set the stuff in the brackets equal to u. For example, if you want to find the integral of 20x(2x + 3)^3, you might then make 2x + 3 = u.

You then need to play with this equation to find two things: an equation for x and an equation for u.

Rearranging the equation, x = (u - 3)/2.
Also, du/dx = 2. Therefore, dx/du = (1/2).

I don't know how to type in integral signs on Blogger, and I don't know what the shorthand is either. So bear with me here.

(integral sign) 20x (2x + 3)^3 dx
= (integral sign) 20((u - 3)/2)(u)^3(dx/du) du
= (integral sign) 10 (u - 3) (u)^3(1/2) du
= (integral sign) (5u - 15)(u)^3 du
= (integral sign) 5u^4 - 15u^3 du
= u^5 - (15/4)u^4 + c
= u^4(u - (15/4)) + c
= (1/4)(u^4)(4u - 15) + c
= (1/4)(2x + 3)^4(8x + 12 - 15) + c
= (1/4)((2x + 3)^4)(8x - 3) + c

You have to make sure that you pop in your substitution for x (in this case x = (u - 3)/2) otherwise you'll end up trying to integrate two variables at once, which I'm told isn't allowed. This is also why you need to have a multiple of the derivative "sitting outside the brackets" in order to integrate. You don't really have to use the substitution method when you do have a multiple of the derivative present, but you still can.

(Well, I'm currently looking at another example which is to find the integral of 8 cos (2x) (sin 2x)^5... but I'm not really sure how to find x in terms of u using the substitution u = sin 2x. At least in this question a multiple of the derivative is present...)