Solving a definite integral without integrating

[Dustinsfl]

You've already correctly said that \sin^4(x)=\frac{1}{4}-\frac{\cos(2x)}{2}+\frac{\cos^2(2x)}{4}, so now you have

\begin{aligned}\int_{-\pi}^{\pi}\sin^4(x)dx &=\int_{-\pi}^{\pi}\left(\frac{1}{4}-\frac{\cos(2x)}{2}+\frac{\cos^2(2x)}{4}\right)dx \\ &= \int_{-\pi}^{\pi}\frac{1}{4}dx-\int_{-\pi}^{\pi}\frac{\cos(2x)}{2}dx+\int_{-\pi}^{\pi}\frac{\cos^2(2x)}{4}dx\end{aligned}

Now, compare these integrals to the inner products that you know. For example, the first one is 1/2 times the inner product of \frac{1}{\sqrt{2}} with itself; which gives you \frac{\pi}{2}...how about the second one?...And the third?

How is the first integral \frac{1}{2}?

 

[gabbagabbahey]

How is the first integral \frac{1}{2}?

It isn't.

\int_{-\pi}^{\pi}\frac{1}{4}dx=\frac{1}{2}\int_{-\pi}^{\pi}\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)dx

It's 1/2 times the inner product of 1/\sqrt{2} with itself.

 

[Dustinsfl]

\frac{1}{2}\int\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{2}}cos(2x)+\frac{1}{4}\int cos(2x)*cos(2x)=\pi*\big[\frac{1}{2}+0+\frac{1}{4}\big]=\frac{3\pi}{4}

I have no idea why the Latex display is all messed up but it is typed in correct.

 

[gabbagabbahey]

You are missing the "dx"s in your integrals, and your integration limits, but other than that it looks fine.

 

[Mark44]

Dustin, I hope you will take what gabbagabbahey and I have been saying to heart.

I understand what you are doing here, but the way you've written it is nonsensical. You've got a grand total of two equal signs in your entire workings; you've got a table without any headings, that looks like a matrix; you've got expressions that are adding integration results to integrands and so on.

I commented on the lack of equal signs much earlier in this thread, and several other points that you didn't acknowledge.

For starters, sin2(x) \neq (1/2)(1 + cos(2x)) [Note: you did fix this one.]

Second, is there any connection between any of the lines above and the one below it? Is there any connection between what you have here and the problem you're trying to solve?

Third, are you trying to convince us that cos(2x) = 1? (see third line) That's not true.

Fourth, is there any connection between the work above and the integral in your first post?

Fifth, is there any connection between your orthogonal basis and what you have here?

What we're trying to get you to do is to write statements (e.g., equations) that flow logically from what you're given to what you want to show. It took a long while to get you to realize the connection between your basis and the integral - i.e., that you were working with something akin to the dot product. Just as the dot product of orthogonal vectors is zero, the inner product of orthogonal functions is also zero.

If it were just me making comments about things, you might put it down to me picking on you, but gabbagabbahey made similar comments about your work. It's in your best interest to keep these comments in mind.