# Thread: a small trigonometric substitution problem.

1. ## a small trigonometric substitution problem.

for the most part, I feel like I understand how to do these. But it breaks down at one very specific point for every problem, and if someone can tell me how to simply approach it, that would be great!

Anyway, heres the problem

integrate (100-x^2)^(1/2) dx from 10-5

In order to save time, I eventually come to100 integrate cos(theta)^2

Then, I find that the integral of that is: (spaces for readability)

(x/2) + sin(2x)/4 +c

The question is, what exactly do I do with this? Im pretty sure I have to draw a right triangle and do something with it. I have the triangle drawn, with the hypotenuse being x, opposite being 10, and the adjacent being the original squared term (100-...)

2. ## Re: a small trigonometric substitution problem.

I take it the question is

$\int^{10}_{5} \frac{1}{\sqrt{100-x^2}} dx$

As

$\int \sqrt{100-x^2} dx = \frac{x}{2} \sqrt{100-x^2} + \frac{100}{2} sin^{-1} (\frac{x}{10}) + C$

??

If it was

$\int \sqrt{1-x^2} dx$

you could sub in $x=sinx$ or $x=cosx$ but I don't think so with 100 in there

3. ## Re: a small trigonometric substitution problem.

Originally Posted by NecroWinter
for the most part, I feel like I understand how to do these. But it breaks down at one very specific point for every problem, and if someone can tell me how to simply approach it, that would be great!

Anyway, heres the problem

integrate (100-x^2)^(1/2) dx from 10-5

In order to save time, I eventually come to100 integrate cos(theta)^2

Then, I find that the integral of that is: (spaces for readability)

(x/2) + sin(2x)/4 +c

The question is, what exactly do I do with this? Im pretty sure I have to draw a right triangle and do something with it. I have the triangle drawn, with the hypotenuse being x, opposite being 10, and the adjacent being the original squared term (100-...)
$I=100 \cdot \int \cos^2 \theta \,d\theta = 50 \cdot \theta + 25 \cdot \sin {2\theta}+C$

Since $~\theta = \arcsin{\frac{x}{10}}~$ it follows that :

$I= \frac{1}{2} \cdot x \cdot \sqrt {100-x^2}+50 \cdot \arcsin{\frac{x}{10}}+C$

4. ## Re: a small trigonometric substitution problem.

Sorry princeps is onto it.

5. ## Re: a small trigonometric substitution problem.

Originally Posted by princeps
$I=100 \cdot \int \cos^2 \theta \,d\theta = 50 \cdot \theta + 25 \cdot \sin {2\theta}+C$

Since $~\theta = \arcsin{\frac{x}{10}}~$ it follows that :

$I= \frac{1}{2} \cdot x \cdot \sqrt {100-x^2}+50 \cdot \arcsin{\frac{x}{10}}+C$
could you be more detailed? i dont know why the square root term is there. i assume you distributed the 100, so im a little lost as to why theres a 1/2 instead of a 50x
when it comes down to it, i think i could use a summary of what you are doing. thanks, and hopefully its not too much trouble, im not asking for a big summary just a few lines really

6. ## Re: a small trigonometric substitution problem.

$\int_5^{10} \sqrt{100-x^2} \, dx$

$x = 10\sin{t}$

$dx = 10\cos{t} \, dt$

substitute and reset the limits of integration ...

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sqrt{100 - 100\sin^2{t}} \cdot 10\cos{t} \, dt$

$100\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sqrt{1 - \sin^2{t}} \cdot \cos{t}\, dt$

$100\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2{t} \, dt$

$50\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 1 + \cos(2t) \, dt$

$50\left[t + \frac{\sin(2t)}{2} \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$

$50\left[\left(\frac{\pi}{2} + 0 \right) - \left(\frac{\pi}{6} + \frac{\sqrt{3}}{4}\right)\right]$

$50 \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$