Integration by parts with substitution

**satis** · September 4th 2011, 01:20 PM

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

The answer says to substitute for $\sqrt{x}$ , so
$u=x^\frac{1}{2}$ .
Thus
$du=\frac{1}{2}x^\frac{-1}{2} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}$

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

$\int{cosy(2ydy)}$

I don't understand this at all... how'd we get from $dx = \frac{1}{2y}dy$ to $cosy(2ydy)$ ? 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.

**Siron** · September 4th 2011, 01:34 PM

If you have:
$\int \cos\left(\sqrt{x}\right)dx$
Let $\sqrt{x}=t \Rightarrow \frac{dx}{2\sqrt{x}}=dt \Leftrightarrow dx=2tdt$
So the integral can be written as:
$\int \cos(t)2tdt$

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

**HallsofIvy** · September 4th 2011, 04:36 PM

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

The answer says to substitute for $\sqrt{x}$ , so
$u=x^\frac{1}{2}$ .
Thus
$du=\frac{1}{2}x^\frac{-1}{2} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}$

This is completely wrong because you forgot the "dx" when you differentiated. You should have
$du= \frac{1}{2}x^{\frac{-1}{2}}dx$
from that, $2x^{\frac{1}{2}}du= dx$ or $2udu= dx$ .

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

$\int{cosy(2ydy)}$

I don't understand this at all... how'd we get from $dx = \frac{1}{2y}dy$ to $cosy(2ydy)$ ? 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.

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