A definite trigonometric absolute integral

[Raghav Gupta]

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?

 

[mfb]

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

 

[Raghav Gupta]

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?

 

[Raghav Gupta]

If n is fraction , then what?

 

[SammyS]

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 $n\pi$ .

 

[Raghav Gupta]

Should we know that n is even or odd?

 

[SammyS]

Should we know that n is even or odd?

Look at a graph, or try some cases.

 

[LCKurtz]

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?

 

[Raghav Gupta]

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

 

[LCKurtz]

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

I don't know. Would it? Have your tried both?

 

[mfb]

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

 

[LCKurtz]

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.

 

[Raghav Gupta]

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.

 

[LCKurtz]

What does $[\cos x]$ mean? Are you saying that you think the antiderivative of $|\sin x|$ is $\cos x$?

 

[mfb]

None of the two is right.

 

[Raghav Gupta]

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.

 

[Raghav Gupta]

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]

I
For odd,
[cosx] from n pi to v,
cosv +1.

The upper limit is not $v$, it is $n\pi + v$.

 

[mfb]

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.

 

[Raghav Gupta]

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 ?

 

[AdityaDev]

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?

 

[AdityaDev]

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.

 

[Raghav Gupta]

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]

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.

 

[SammyS]

Split your integral as follows:
$\displaystyle\ \int_0^{nπ+v} |sinx| dx = \int_0^{v} |sinx| dx + \int_v^{nπ+v} |sinx| dx$
and apply what Zondrina recently posted.

 

[Raghav Gupta]

Thank you guys, once again.