# Thread: Strange integration by substitution

1. ## Strange integration by substitution

Hi there,

I've been given a question to do as homework, it seems pretty weird to me:

Use the substitution x = 4cosθ to evaluate the following definite integral:

$\displaystyle \int^2_0\frac{x}{16-x^2} dx$
Any help will be greatly appreciated.

Thank you.

2. Originally Posted by title
Strange integration by substitution
I would paraphrase that to:
Originally Posted by title
Integration by strange substitution
$\displaystyle \int^2_0\frac{x}{16-x^2} dx$

Using the sub $\displaystyle x = 4\cos{\theta} \implies dx = -4\sin{\theta}~d\theta$ yields:

$\displaystyle -16 \int \frac{\sin{\theta}\cos{\theta}}{16(1-\cos^2{\theta})}~d\theta = -\int \frac{\cos{\theta}}{\sin{\theta}}~d\theta = -\ln{|\sin{\theta}|} + C$

Now back substitute and evaluate

3. Originally Posted by Chop Suey
I would paraphrase that to:

$\displaystyle \int^2_0\frac{x}{16-x^2} dx$

Using the sub $\displaystyle x = 4\cos{\theta} \implies dx = -4\sin{\theta}~d\theta$ yields:

$\displaystyle -16 \int \frac{\sin{\theta}\cos{\theta}}{16(1-\cos^2{\theta})}~d\theta = -\int \frac{\cos{\theta}}{\sin{\theta}}~d\theta = -\ln{|\sin{\theta}|} + C$

Now back substitute and evaluate
The substitution $\displaystyle u = 16 - x^2$ makes for an easier calculation.

4. Originally Posted by mr fantastic
The substitution $\displaystyle u = 16 - x^2$ makes for an easier calculation.
Indeed, but the question asked to use x = 4cosθ, so...

5. Originally Posted by Chop Suey
Indeed, but the question asked to use x = 4cosθ, so...
Indeed. A pity the question didn't also ask that I wear glasses while reading it.