# Substitution Rule in Indefinite Integrals

## Homework Statement

Please explain how to use the substitution rule in indefinite integrals. I am unable even to start the problem.

## The Attempt at a Solution

#### Attachments

• 1.7 KB Views: 424

Actually, one second. I am solving it right now. Will post a solution.

Mark44
Mentor
Substitution is the reverse of the chain rule in differentiation. For example,
$$\frac{d}{dx}(x^3 - 2)^4~=~4(x^3 - 2)^3~3x^2$$

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

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

Thank you for the helpful note. I did not know that substitution is reverse of chain rule. Here is my solution. Correct?

#### Attachments

• 9.4 KB Views: 403
Mark44
Mentor
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.

Why do you put du=2xdx? I think we need dx=du/2x.

Here is the completed version.

#### Attachments

• 10.4 KB Views: 394
Can you check another problem please too? Thank you.

#### Attachments

• 8.3 KB Views: 428
Mark44
Mentor
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

$$\int sec(2t)tan(2t)dt = \frac{1}{2}\int 2sec(2t)tan(2t)dt = \frac{1}{2}\int du= \frac{u}{2} + C = \frac{sec(2t)}{2} + C$$

Thank you very much!

I am also stuck on this one. ex has been always tricky for me.

#### Attachments

• 5.7 KB Views: 392
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

$$\int sec(2t)tan(2t)dt = \frac{1}{2}\int 2sec(2t)tan(2t)dt = \frac{1}{2}\int du= \frac{u}{2} + C = \frac{sec(2t)}{2} + C$$

OH! This one is much prettier than mine!

Mark44
Mentor
Thank you very much!

I am also stuck on this one. ex has been always tricky for me.
This will work, but will require another substitution.

Here's a different approach: u = ex + 1, du = exdx
$$\int \frac{e^x~dx}{e^x + 1}=\int \frac{du}{u}= ln(u) + C = ln(e^x+ 1) + C$$

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

Thank you a lot, Mark44!