Homework Help: Substitution Rule in Indefinite Integrals

Substitution is the reverse of the chain rule in differentiation. For example,
[tex]\frac{d}{dx}(x^3 - 2)^4~=~4(x^3 - 2)^3~3x^2[/tex]

The corresponding indefinite integral is
[tex]\int 12x^2(x^3 - 2)^3~dx[/tex]
Here, we let u = x³ - 2, so du = 3x²~dx, so
[tex]\int 12x^2(x^3 - 2)^3~dx~=~\int 4*3x^2~(x^3 - 2)^3~dx~=~\int 4 u^3 du~=~\frac{4u^4}{4} + C~=~(x^3 - 2)^4 + C[/tex]

Can you see how this technique (it's not a rule) could be applied to your problem?

 

Correct as far as it goes. You still need to undo your substitution.

Also, u = x² + 1 ==> du = 2xdx. You have x and dx, so you can get the 2 that's needed by multiplying by 2 and then multiplying on the outside of the integral by 1/2. That way, you're really multiplying by 1, which doesn't change the value of the integrand.

 

Sure, that works.

Another substitution that isn't so obvious is to let u = sec2t (using t for theta). Then du = 2sec(2t)tan(2t)dt

[tex]\int sec(2t)tan(2t)dt = \frac{1}{2}\int 2sec(2t)tan(2t)dt = \frac{1}{2}\int du= \frac{u}{2} + C = \frac{sec(2t)}{2} + C[/tex]

 

This will work, but will require another substitution.

Here's a different approach: u = e^x + 1, du = e^xdx
[tex]\int \frac{e^x~dx}{e^x + 1}=\int \frac{du}{u}= ln(u) + C = ln(e^x+ 1) + C[/tex]

Since u = e^x + 1 > 0 for all real x, I didn't need to include absolute values on the argument to the natural log function.