Find the result of definite integration

[songoku]

Homework Statement

Find $\frac{2}{L} \int_{0}^{L} \sin\left(\frac{5x}{L}\right) \sin \left(\frac{n\pi x}{L} \right) dx$

Relevant Equations

Trigonometry Identity

Integration

$$\frac{2}{L} \int_{0}^{L} \sin\left(\frac{5x}{L}\right) \sin \left(\frac{n\pi x}{L} \right)dx$$
$$=\frac{2}{L} \int_{0}^{L} \left(\frac{1}{2} \cos\left[\left(\frac{5-n\pi}{L}\right)x\right] -\frac{1}{2}\cos \left[\left(\frac{5 + n\pi}{L} \right)x\right]\right)dx$$
$$=\frac{1}{5-n\pi} \sin (5-n\pi)-\frac{1}{5+n\pi}\sin(5+n\pi)$$

This is as far as I can get but the answer key is if $n=5$ then the result of integration is 1 and will be zero for other values of $n$.

How to proceed to get the answer? Thanks

 

[renormalize]

Homework Statement: Find $\frac{2}{L} \int_{0}^{L} \sin\left(\frac{5x}{L}\right) \sin \left(\frac{n\pi x}{L} \right) dx$

Are you sure there's not a factor of $\pi$ missing in that expression? I think it should read:$$\frac{2}{L} \int_{0}^{L} \sin\left(\frac{5\pi x}{L}\right) \sin \left(\frac{n\pi x}{L} \right) dx$$

 

[songoku]

Are you sure there's not a factor of $\pi$ missing in that expression? I think it should read:$$\frac{2}{L} \int_{0}^{L} \sin\left(\frac{5\pi x}{L}\right) \sin \left(\frac{n\pi x}{L} \right) dx$$

No, in the original question there is no $\pi$

But let say if there is $\pi$, then the integration becomes:
$$=\frac{1}{5\pi-n\pi} \sin (5\pi-n\pi)-\frac{1}{5\pi+n\pi}\sin(5\pi+n\pi)$$

In this case, $n$ can not be 5 because one of the denominator will become 0 and for other value of $n$ (sorry I forgot to put that $n$ must be positive integer) the sin term will be zero. Still does not match the answer key for $n=5$

Thanks

EDIT: I just realize the numerator of the first term is also zero. Does it mean I have to use limit as $n$ approaches 5 to find the value of the first term? I tried and the result is 1

 

[Gavran]

EDIT: I just realize the numerator of the first term is also zero. Does it mean I have to use limit as $n$ approaches 5 to find the value of the first term? I tried and the result is 1

Correct.

 

[renormalize]

No, in the original question there is no $\pi$

Well then there's definitely a typo in the original homework statement. And yes, your limit approach gives the correct answer for $n=5$. But you can also get the correct result by setting $n=5$ in the integrand before you do the integral. Try it!

 

[PeroK]

But let say if there is $\pi$, then the integration becomes:
$$=\frac{1}{5\pi-n\pi} \sin (5\pi-n\pi)-\frac{1}{5\pi+n\pi}\sin(5\pi+n\pi)$$

When you divide by something, you must always specify that something is not zero. For example:
$$\int \cos(nx) \ dx = \frac 1 n \sin(nx) + C \ (n \ne 0)$$In this case, if $n = 0$, then:
$$\int \cos(nx) \ dx = \int 1 \ dx = x + C$$

 

[songoku]

Well then there's definitely a typo in the original homework statement. And yes, your limit approach gives the correct answer for $n=5$. But you can also get the correct result by setting $n=5$ in the integrand before you do the integral. Try it!

When you divide by something, you must always specify that something is not zero. For example:
$$\int \cos(nx) \ dx = \frac 1 n \sin(nx) + C \ (n \ne 0)$$In this case, if $n = 0$, then:
$$\int \cos(nx) \ dx = \int 1 \ dx = x + C$$

Ah I see. Since I need to divide by $5\pi-n\pi$, I need to realize this is for the case where $n\neq 5$. The case $n=5$ must be done separately.

Thank you very much for the help and explanation renormalize, Gavran, PeroK