Positive sqrt of trig functions

Stupid question.

In some integration problems, I've found myself taking the positive root of some functions (usually in substitution).

Why is it possible to do so?

For example, I might say

sinx=\sqrt{1-cos^2x}

.
I have a bit of an idea.

Reply 1

user2020user

16

For definite integrals, we'll have to consider the limits of the integration and whether the modulus of that function is positive or negative. It's in a box in my solution to your integral on the A-Level Maths Thread MK III.

As for indefinite integrals, beats me. I guess you'd need to set up a piecewise defined answer.

(edited 10 years ago)

Reply 2

ztibor

10

Original post by keromedic

Stupid question.

In some integration problems, I've found myself taking the positive root of some functions (usually in substitution).

Why is it possible to do so?

For example, I might say

sinx=\sqrt{1-cos^2x}

.
I have a bit of an idea.

As the

1-\cos^2 x

is nonnegative real it has a unique non-negative square root, called the principal square root, which is denoted by

\sqrt{1-\cos^2 x}

So this means nonegative value by definition.
For other root use

-\sqrt{1-\cos^2 x}

denotion

Reply 3

davros

16

Original post by keromedic

Stupid question.

In some integration problems, I've found myself taking the positive root of some functions (usually in substitution).

Why is it possible to do so?

For example, I might say

sinx=\sqrt{1-cos^2x}

.
I have a bit of an idea.

You need to be very careful when doing this!

The square root sign by definition indicates the positive square root when applied to a positive real number.

However, sin x can be positive or negative depending on the value of x, so although it is always true that

\sin^2 x = 1 - \cos^2 x

you should note that

\sqrt{\sin^2 x} = |\sin x|

and this may not be the same as sin x!

Post a specific example if you need reassurance about what you've done in an integral.

Reply 4

user2020user

16

Original post by davros

Post a specific example if you need reassurance about what you've done in an integral.

I was thinking along the same lines when I did this question:

Original post by keromedic

\displaystyle \int_0^{\frac{\pi}{4}} \sin (x) \sqrt{\cos (2x)} \text{ d}x

Solution:

Spoiler

How would we have dealt with the integral had there not been any lower or upper limits to it?

Could we, as I mentioned very briefly before, define the final result in a piece-wise manner for each of the regions where

\theta

satisfies the following?

|\tan \theta| = + \tan \theta

| \tan \theta | = -\tan \theta

(edited 10 years ago)

Reply 5

davros

16

Original post by Khallil

I was thinking along the same lines when I did this question:

Solution:

Spoiler

How would we have dealt with the integral had there not been any lower or upper limits to it?

Could we, as I mentioned very briefly before, define the final result in a piece-wise manner for each of the regions where

\theta

satisfies the following?

|\tan \theta| = + \tan \theta

| \tan \theta | = -\tan \theta

My immediate reaction (and this probably needs more thought than I'm giving it, as I'm a bit sleep-deprived today!) is that you need to be careful trying to write down a general form for the indefinite integral because there are clearly going to be values of x where cos(2x) is negative and hence the integrand won't even exist!

I'd suggest you play around with it for a bit - maybe graph the integrand itself to get an idea what it looks like in different regions and then see where the integral does and doesn't make sense! I would hope in a normal exam situation issues like this don't arise but you're right to be asking questions about it :smile:

Positive sqrt of trig functions

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