# Thread: Integration via substitution

1. ## Integration via substitution

Hi all,

I've got a bit of a problem with this question, if anyone could show the working it would be appreciated. I've got the suspicion it's meant to be done via substitution, but looking to learn.

$\int^{\pi/8}_0 (\theta + \cos4\theta) d\theta$

I'm pretty sure that the answer is (32 + pi^2) / 128 but of course, the answer isn't everything.

Anyone able to help me out? Cheers.

2. Originally Posted by Peleus
Hi all,

I've got a bit of a problem with this question, if anyone could show the working it would be appreciated. I've got the suspicion it's meant to be done via substitution, but looking to learn.

$\int^{\pi/8}_0 (\theta + \cos4\theta) d\theta$

I'm pretty sure that the answer is (32 + pi^2) / 128 but of course, the answer isn't everything.

Anyone able to help me out? Cheers.
Break it up into the sum of two integrals. One can be done directly, the other via a simple substitution.

3. Here's my working so far, however I think I've done something wrong.

$
\int^{\pi/8}_0 (\theta + \cos4\theta) d\theta
$

$
\int^{\pi/8}_0 \theta d\theta + \int^{\pi/8}_0 \cos4\theta d\theta
$

$
\int^{\pi/8}_0 \theta d\theta + \cos4\int^{\pi/8}_0 \theta d\theta
$

$
[\frac{\theta^2}{2}]^{\pi/8}_0 + [\cos2\theta^2]^{\pi/8}_0
$

4. It's cos(4theta), presumably, not cos(4)*theta. You can't pull it out of the integral.

5. Originally Posted by Peleus
[snip]
$
\int^{\pi/8}_0 \theta d\theta + \cos4\int^{\pi/8}_0 \theta d\theta
$

[snip]
And would you also say that $\int \sqrt{x} \, dx = \sqrt{{\color{white} ..}} \int x \, dx$. I sincerely hope not! You need to thoroughly review functions as a matter of urgency.

6. Yeah looking back it's quite silly of me. Brain snap.

This could be just as silly, but I think it's better.

Let $4\theta = x$ and $dx = 4d\theta$

$
\int^{\pi/8}_0 \cos x dx
$

$
\int^{\pi/8}_0 \frac{1}{4} \sin x
$

$
[\frac{1}{4} \sin(4\theta)]^{\pi/8}_0
$

$
= \frac{1}{4}
$

7. Originally Posted by Peleus
Yeah looking back it's quite silly of me. Brain snap.

This could be just as silly, but I think it's better.

Let $4\theta = x$ and $dx = 4d\theta$

$
\int^{\pi/8}_0 \cos x dx
$

$
\int^{\pi/8}_0 \frac{1}{4} \sin x
$

$
[\frac{1}{4} \sin(4\theta)]^{\pi/8}_0
$

$
= \frac{1}{4}
$
No need to go back to the original variable with definite integrals. Just substitute in the integral terminals as well. $\theta = 0 \Rightarrow x = ....$ and $\theta = \frac{\pi}{8} \Rightarrow x = ....$ and evaluate the new definite integral. Saves time and money.