# Trigonometric Integrals

1. Jan 25, 2004

### 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

2. Jan 25, 2004

### Hurkyl

Staff Emeritus
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:

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

3. Jan 25, 2004

### noboost4you

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.

I got nothing. Not even a hunch.

4. Jan 25, 2004

### Hurkyl

Staff Emeritus
1) Do you remember integration by parts?

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

5. Jan 25, 2004

### noboost4you

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?

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

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

6. Jan 25, 2004

### 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?

7. Jan 25, 2004

### Hurkyl

Staff Emeritus
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?)

8. Jan 25, 2004

### noboost4you

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

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

9. Jan 25, 2004

### noboost4you

They all equal zero

10. Jan 25, 2004

### PrudensOptimus

&int;xcos^2x dx = x[0.5(x + sin2x/2)] - &int;[0.5(x + sin2x/2)]dx

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

11. Jan 25, 2004

### Hurkyl

Staff Emeritus
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?

12. Jan 26, 2004

### 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