Integration Problem: Solving \int^{0}_{-\pi}\sqrt{1-cos^{2} x} for Homework

[ngkamsengpeter]

Homework Statement

$$\int^{0}_{-\pi}\sqrt{1-cos^{2} x}$$

Homework Equations

The Attempt at a Solution

I substitute into (sin x)^2 and get an answer of -2 but the answer should be 2 . how to i do this question.

 

[cristo]

How do you get an answer of -2? Perhaps you should show your work.

 

[ngkamsengpeter]

How do you get an answer of -2? Perhaps you should show your work.

just substitute (sin x)^2 and become sqr((sin x)^2) and then become sin x and substitute the limit from -pi to 0 get -2

 

[cristo]

just substitute (sin x)^2 and become sqr((sin x)^2) and then become sin x and substitute the limit from -pi to 0 get -2

You need to integrate sin(x) before you plug in the limits.

 

[ngkamsengpeter]

You need to integrate sin(x) before you plug in the limits.

I have integrate it into -cos x and plugin the limit , i got -2 .But the answer is 2 . How ?

 

[cristo]

So you have $$\left[-\cos(x)\right]^0_{-\pi}=cos(0)-cos(-\pi)$$. Can you evaluate that?

 

[jpr0]

maybe you should be interpreting $\sqrt{\sin^2(x)}$ as $|\sin(x)|$

 

[sutupidmath]

So you have $$\left[-\cos(x)\right]^0_{-\pi}=cos(0)-cos(-\pi)$$. Can you evaluate that?[/QUOTE

$$\left[-\cos(x)\right]^0_{-\pi}=-cos(0)+cos(-\pi)$$
shouldn't is be this way cristo.

 

[ngkamsengpeter]

maybe you should be interpreting $\sqrt{\sin^2(x)}$ as $|\sin(x)|$

Then , how to integrate $|\sin(x)|$

 

[HallsofIvy]

For x between $-\pi$ and 0, sin(x)< 0. In that range, |sin(x)| is just -sin(x). Integrating that will obviously give you the negative of your previous answer.

 

[transgalactic]

but ((sin x)^2 )^0.5 gives us two answers

sinx and -sinx

?

 

[sutupidmath]

but ((sin x)^2 )^0.5 gives us two answers

sinx and -sinx

?

yeah, generally it does, but look here we are only integrating in [-pi.0], and obviously sinx, where x is from [-pi,0] is always negative, so

I sin(x) I = -sinx, whenever x is from the interval [-pi. 0]
now as halls said, integrating this you will get the desired answer.

 

[ngkamsengpeter]

yeah, generally it does, but look here we are only integrating in [-pi.0], and obviously sinx, where x is from [-pi,0] is always negative, so

I sin(x) I = -sinx, whenever x is from the interval [-pi. 0]
now as halls said, integrating this you will get the desired answer.

But i use mathematica to integrate , it shows -Cot x ((Sin x)^2)^(1/2)
How to integrate to get this form ?

 

[HallsofIvy]

Why would you want to? This is a definite integral. The result is a number. The integrand reduces to |sin(x)| which, for $-\pi\le x\le 0$ is -sin(x). That's easy to integrate.

I've never used mathematica and what you give makes me glad I haven't! It's clearly using some general algorithm and then not recognizing that, since $(sin^2(x))^(1/2)$ is |sin(x)|, $-cot(x)(sin^2(x))^(1/2)= -(cos(x)/sin(x))(-sin(x))= cos(x)$.