Integrals of x cos²x and sin¹³⁷x from -π to π

[noboost4you]

I've been able to do all of them up until these two:

1) $$\int x\cos^{2}x dx$$

and

2) $$\int$$ from $$-\pi$$ to $$\pi$$ of $$\sin^{137}x dx$$

I've been using half angle formulas where (cos x)^2 = (1/2)(1 + cos 2x) and (sin x)^2 = (1/2)(1 - cos 2x)

I just can't figure those two out. Any help would be highly appreciated

 

[Hurkyl]

Here's some hints:

1) Do you know how to integrate $\int x e^x \, dx$?

2) A large number (like 137) is often a hint that there's something more interesting at stake. Try some smaller powers and see if you can recognize a pattern, and then prove it.


P.S. if you were wondering how to make the integral, it's this:

$$\int_{-\pi}^{\pi} \sin^{137} x \, dx$$

 

[noboost4you]

Originally posted by Hurkyl
Here's some hints:

1) Do you know how to integrate $\int x e^x \, dx$?

If this was last semester, yes, but now I can't remember. I know $$\int e^x \, dx$$ equals $$e^x + c$$. Substitution plays a role, but I just can't see how.

2) A large number (like 137) is often a hint that there's something more interesting at stake. Try some smaller powers and see if you can recognize a pattern, and then prove it.

I got nothing. Not even a hunch.

 

[Hurkyl]

1) Do you remember integration by parts?


2) So what smaller powers have you tried and what were the values of the integrals?

 

[noboost4you]

Originally posted by Hurkyl
1) Do you remember integration by parts?

Yes, let me try to solve your equation now.

$\int x e^x \, dx$

u = $$x$$ du = $$dx$$
dv = $$e^x$$ v = $$e^x$$

Therefore $\int x e^x \, dx$ = $$x e^x - \int e^x dx$$ which in turn equals $$x e^x - 2x e^x + 2x + c$$

Correct?

2) So what smaller powers have you tried and what were the values of the integrals?

$$\int_{-\pi}^{\pi} \sin^{137} x \, dx$$ = $$\int_{-\pi}^{\pi} \sin^{135} x \sin^{2} x \, dx$$ and $$\sin^{2} x = 1/2(1 - \cos 2x)$$

I'm slowly getting it. Thanks for the help as far.

 

[noboost4you]

Back to #1 $$\int x\cos^{2}x dx$$

u = $$x$$ du = $$dx$$
dv = $$\cos^{2} x$$ v = $$\sin^{2} x$$

= $$x \sin^{2} x - \int \sin^{2} x dx$$
= $$x \sin^{2} x - (1/2)\int (1 - \cos 2x) dx$$
= $$x \sin^{2} x - (x/2) - ( \sin 2x / 4) + c$$

how's that look?

 

[Hurkyl]

Well, $x e^x - \int e^x \, dx = x e^x - e^x + C$, but I think you knew that and just made a typo. And you should really always include the $dx$ part, so you say that $dv = e^x \, dx$.


Now, when you attacked your actual problem, you have $dv = \cos^2 x \, dx$. You got $v = sin^2 x$, but that doesn't work because $dv = 2 \sin x \cos x \, dx$! Remember this says that the derivative of $v$ has to be $\cos^2 x$, so we need to integrate to find $v$! In other words:

$$v = \int \cos^2 x \, dx$$



For problem 2:

There is a really "obvious" method for this problem that takes advantage of the shape of your integrand and the domain of integration... one you'll recognize with a little practice. I'm trying not to spoil the exercise by just giving you the answer, which is why I'm suggesting to try some small cases; I want you to find:

$$\int_{-\pi}^{\pi} \sin^1 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^2 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^3 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^4 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^5 x \, dx$$

That should probably be enough to see the relevant pattern.


(Just to check, you have learned how to integrate odd powers of the sine function right?)

 

[noboost4you]

Originally posted by Hurkyl
Well, $x e^x - \int e^x \, dx = x e^x - e^x + C$, but I think you knew that and just made a typo. And you should really always include the $dx$ part, so you say that $dv = e^x \, dx$.


Now, when you attacked your actual problem, you have $dv = \cos^2 x \, dx$. You got $v = sin^2 x$, but that doesn't work because $dv = 2 \sin x \cos x \, dx$! Remember this says that the derivative of $v$ has to be $\cos^2 x$, so we need to integrate to find $v$! In other words:

$$v = \int \cos^2 x \, dx$$

$$v = \int \cos^2 x \, dx$$ = $$x/2 + \sin 2x/4$$ ??

For problem 2:

There is a really "obvious" method for this problem that takes advantage of the shape of your integrand and the domain of integration... one you'll recognize with a little practice. I'm trying not to spoil the exercise by just giving you the answer, which is why I'm suggesting to try some small cases; I want you to find:

$$\int_{-\pi}^{\pi} \sin^1 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^2 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^3 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^4 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^5 x \, dx$$

That should probably be enough to see the relevant pattern.


(Just to check, you have learned how to integrate odd powers of the sine function right?)

Basically what you're saying is that no matter the number sine is raised to, the integrand between $$-\pi$$ and $$\pi$$ will always be the same?? Therefore I should just use $$\int_{-\pi}^{\pi} \sin^2 x \, dx$$ and solve that? Yes? Maybe? Not even close? Slowly, but surely

 

[noboost4you]

Originally posted by Hurkyl


For problem 2:

There is a really "obvious" method for this problem that takes advantage of the shape of your integrand and the domain of integration... one you'll recognize with a little practice. I'm trying not to spoil the exercise by just giving you the answer, which is why I'm suggesting to try some small cases; I want you to find:

$$\int_{-\pi}^{\pi} \sin^1 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^2 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^3 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^4 x \, dx$$
$$\int_{-\pi}^{\pi} \sin^5 x \, dx$$

That should probably be enough to see the relevant pattern.

They all equal zero

 

[PrudensOptimus]

∫xcos^2x dx = x[0.5(x + sin2x/2)] - ∫[0.5(x + sin2x/2)]dx

= ... - (x^2/4 - cos[2x]/8)

 

[Hurkyl]

Yyou got v right.


Actually, only the odd powers of sin turn out to integrate to zero (and 137 is an odd power); can you figure out why?

 

[himanshu121]

Simply it is an odd function and they are symmetrical w.r.t Origin or in (I &III) coordinate hence one portion will be above x-axis and other below x-axis with equal magnitude hence the result would be zero