# A definite trigonometric absolute integral

## Homework Statement

Find ## \int_0^{nπ+v} |sinx| dx ##
Options are,
2n+ 1+ cosv
2n+1-cosv
2n+1
2n+cosv

## Homework Equations

Integration of sinx is -cosx.

## The Attempt at a Solution

Sin x is + ve from 0 to π,
Negative from π to 2π
We can make |sinx| as sinx in 0 to π,
And - sinx in π to 2π
But here n is confusing.
We will rotate in a cycle?

## Answers and Replies

mfb
Mentor
What is the integral up to pi, up to 2pi, 3pi, ... n pi?
What is the integral from n pi to n pi + v?

Upto pi it is 2, upto 2 pi it is 4.
So for upto n pi it is 2n.
But for integral from n pi to n pi +v , we should know whether n is even or odd, isn't it?

If n is fraction , then what?

SammyS
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If n is fraction , then what?
It's pretty clear that ##n## is intended to be a natural number. ##v## takes care of any extra fractional part beyond ##n\pi## .

Should we know that n is even or odd?

SammyS
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Should we know that n is even or odd?
Look at a graph, or try some cases.

LCKurtz
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Should we know that n is even or odd?

Does it matter in the answer? And I don't see where you stated it anywhere but I guess we are assuming ##0\le v <\pi##, right?

Wouldn't the sign change in modulus function for even pi?

LCKurtz
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Wouldn't the sign change in modulus function for even pi?
I don't know. Would it? Have your tried both?

mfb
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The sign of the sine changes, but the sign of |sin(x)| does not.

LCKurtz
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The sign of the sine changes, but the sign of |sin(x)| does not.
...as both SammyS and I have been trying to get the OP to discover for himself.

• SammyS
I'm getting diff. Answers considering even and odd.
For n to be even,
-[ cosx ] from n pi to v,
It's
-(cosv - 1)= 1-cosv
For odd,
[cosx] from n pi to v,
cosv +1.

Last edited:
LCKurtz
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What does ##[\cos x]## mean? Are you saying that you think the antiderivative of ##|\sin x|## is ##\cos x##?

mfb
Mentor
None of the two is right.

What does ##[\cos x]## mean? Are you saying that you think the antiderivative of ##|\sin x|## is ##\cos x##?
No ant iderivative of sin x is -cosx and of -sinx is cosx.

None of the two is right.
I see that there was a time lapse of 1 min between my editing of 13th post and your reply.
Can you reconsider it?
Regards.

LCKurtz
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I
For odd,
[cosx] from n pi to v,
cosv +1.
The upper limit is not ##v##, it is ##n\pi + v##.

mfb
Mentor
The problem is exactly the same for odd and even n, because in both cases v is between 0 and pi and the sine goes from 0 to 1 and back to 0 again. I don't see why you make two categories.

See this,
## \int_0^{nπ+v} |sinx| dx ##

= ## \int_0^π sinx dx - \int_π^{2π} sinx dx + \int_{2π}^{3π} sinx dx + .... \int_{nπ}^{v} sinx dx ##
= ##2n + \int_{nπ}^{v} sinx dx ##
= ## 2n - (cosx) ## here cos x has limits nπ to v, ( I don't know the latex for that).
= ## 2n - (cosv - cosnπ) ##
Now surely,here if n is even, odd it matters, cos value being -1 for odd π and +1 for even π ?

Got it... its not np to v. Its np to np+v.
$$2n+\int_{n\pi}^{n\pi+v}|sinx|dx$$
Now apply property ##\int_{nT}^{nT+a}f(x)dx=\int_0^af(x)dx##
Since ##|sinx|## has a period of π.
Now your integral becomes very easy:
$$I=2n+\int_0^vsinxdx$$

• Raghav Gupta
Thanks very much Aditya. I don't know that time period property. Can you give a link or reference for the proof of it?

Its intuitive. Draw a graph of sinx and shade the area from p to p+2 and shade the area between 0 to 2. You have to prove it using graph.

Okay got the intuitive idea. I had to see Khan Academy video to clear this graph thing. Thanks.

Also thanks to other guys for giving me a approach to solve problem.

STEMucator
Homework Helper
Thanks very much Aditya. I don't know that time period property. Can you give a link or reference for the proof of it?

Suppose that ##f## is a ##T## periodic function. Then for any real number ##\alpha##, we can write:

$$\int_{\alpha}^{\alpha + T} f(x) \space dx = \int_0^T f(x) \space dx$$

To prove this, write:

$$\int_{\alpha}^{\alpha + T} f(x) \space dx = \int_{\alpha}^0 f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx$$

Now suppose ##u = x + T \Rightarrow x = u - T \Rightarrow dx = du## for the first integral on the right, then:

$$\int_{\alpha}^0 f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(u - T) \space du + \int_{0}^{\alpha + T} f(x) \space dx$$

Now due to the ##T## periodicity of ##f##, we may write:

$$\int_{\alpha + T}^{T} f(u - T) \space du + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(u) \space du + \int_{0}^{\alpha + T} f(x) \space dx$$

Simply replace the dummy variable ##u## by ##x## now to obtain:

$$\int_{\alpha + T}^{T} f(u) \space du + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx$$

The integrals represent the area from ##[0, T]## because we integrate from ##0## to ##\alpha + T## and then from ##\alpha + T## to ##T##. Therefore:

$$\int_{\alpha + T}^{T} f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx = \int_0^T f(x) \space dx$$

So really, the integral of a ##T## periodic function over any interval of length ##T## is the same.

• Raghav Gupta