Integration by parts with substitution

Integration by parts I understand, but part of the problems for this section has instructions to "First make a substitution and then use integration by parts to evaluate the integral". The book itself doesn't even mention this in the section. I have the answers, but it's not really helping me understand what's going on or why.

example:

$\displaystyle \int{cos\sqrt{x}dx}$

The answer says to substitute for $\displaystyle \sqrt{x}$, so

$\displaystyle u=x^\frac{1}{2}$.

Thus

$\displaystyle du=\frac{1}{2}x^\frac{-1}{2} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}$

Switching in the y is a bit of a stretch. I understand it, but I'm not sure I understand why. The book then shows to substitute all this back into the original problem to get

$\displaystyle \int{cosy(2ydy)}$

I don't understand this at all... how'd we get from $\displaystyle dx = \frac{1}{2y}dy$ to $\displaystyle cosy(2ydy)$? From there, it's just integration by parts, and I understand the process. It's only the substitution part that's got me baffled.

If someone is able to explain the how and why of that, I'd be much appreciative.

Re: Integration by parts with substitution

If you have:

$\displaystyle \int \cos\left(\sqrt{x}\right)dx$

Let $\displaystyle \sqrt{x}=t \Rightarrow \frac{dx}{2\sqrt{x}}=dt \Leftrightarrow dx=2tdt$

So the integral can be written as:

$\displaystyle \int \cos(t)2tdt$

Note:

If you use the variable $\displaystyle u$ then it's confusing to suddenly use the variable $\displaystyle y$ so stay with one variable.

Re: Integration by parts with substitution

Quote:

Originally Posted by

**satis** Integration by parts I understand, but part of the problems for this section has instructions to "First make a substitution and then use integration by parts to evaluate the integral". The book itself doesn't even mention this in the section. I have the answers, but it's not really helping me understand what's going on or why.

example:

$\displaystyle \int{cos\sqrt{x}dx}$

The answer says to substitute for $\displaystyle \sqrt{x}$, so

$\displaystyle u=x^\frac{1}{2}$.

Thus

$\displaystyle du=\frac{1}{2}x^\frac{-1}{2} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}$

This is completely wrong because you forgot the "dx" when you differentiated. You should have

$\displaystyle du= \frac{1}{2}x^{\frac{-1}{2}}dx$

from that, $\displaystyle 2x^{\frac{1}{2}}du= dx$ or $\displaystyle 2udu= dx$.

Quote:

Switching in the y is a bit of a stretch. I understand it, but I'm not sure I understand why. The book then shows to substitute all this back into the original problem to get

$\displaystyle \int{cosy(2ydy)}$

I don't understand this at all... how'd we get from $\displaystyle dx = \frac{1}{2y}dy$ to $\displaystyle cosy(2ydy)$? From there, it's just integration by parts, and I understand the process. It's only the substitution part that's got me baffled.

If someone is able to explain the how and why of that, I'd be much appreciative.