A definite trigonometric absolute integral

1. Mar 29, 2015

Raghav Gupta

1. The problem statement, all variables and given/known data

Find $\int_0^{nπ+v} |sinx| dx$
Options are,
2n+ 1+ cosv
2n+1-cosv
2n+1
2n+cosv
2. Relevant equations
Integration of sinx is -cosx.

3. 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?

2. Mar 29, 2015

Staff: 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?

3. Mar 29, 2015

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?

4. Mar 29, 2015

Raghav Gupta

If n is fraction , then what?

5. Mar 29, 2015

SammyS

Staff Emeritus
It's pretty clear that $n$ is intended to be a natural number. $v$ takes care of any extra fractional part beyond $n\pi$ .

6. Mar 29, 2015

Raghav Gupta

Should we know that n is even or odd?

7. Mar 29, 2015

SammyS

Staff Emeritus
Look at a graph, or try some cases.

8. Mar 29, 2015

LCKurtz

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?

9. Mar 29, 2015

Raghav Gupta

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

10. Mar 29, 2015

LCKurtz

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

11. Mar 29, 2015

Staff: Mentor

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

12. Mar 29, 2015

LCKurtz

...as both SammyS and I have been trying to get the OP to discover for himself.

13. Mar 29, 2015

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.

Last edited: Mar 29, 2015
14. Mar 29, 2015

LCKurtz

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

15. Mar 29, 2015

Staff: Mentor

None of the two is right.

16. Mar 29, 2015

Raghav Gupta

No ant iderivative of sin x is -cosx and of -sinx is cosx.

17. Mar 29, 2015

Raghav Gupta

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.

18. Mar 29, 2015

LCKurtz

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

19. Mar 29, 2015

Staff: 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.

20. Mar 30, 2015

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 ?