# Math-inclined friends, I need your help!

So we’re doing trigonometric integrals in calculus and one of our homework problems is this little dude: We rushed through this section of the chapter this morning ’cause we’re behind schedule and I’m a little shaky on them (also I’m dumb), so I went to the calc room in Polya to get some help. I showed one of the tutors in there this integral.

I told him how I thought we should start: since (1-cos2x) is the numerator of the half-angle formula for sin2x, we could multiply both sides of the half-angle formula to change (1-cos2x) to 2sin2x and then go from there.

He said he’d never even thought about solving it like that, but when I asked him what the “normal” method would be for this integral, he didn’t know.

So is there another way of solving this?

### 2 responses

1. I’d do pretty much exactly what you’re thinking. I came by the 2sinsquared(x) a little bit differently (sorry, no idea how to do superscripts in a WP comment box…) but the end result is the same. I substituted 1-2sinsquared(x) for cos(2x) via the double angle formulas for cosine. Then of course the 1’s cancel out and you’re left with just 2sinsquared(x) same as in your method. Then applied the exponent surrounding it, took the constant 8*2^(1/2) out of the integral which leaves just sincubed(x)dx inside, replaced sincubed(x) with (sin(x))(1-cossquared(x)), and multiplied these to get sin(x)-sin(x)cossquared(x). Then separated this into 2 integrals, but left -1 inside the second integral to make substitution easier. Then substituted u=cos(x) and du=-sin(x)dx in the second integral, and from there it’s pretty straightforward.

Ugh…that was totally incoherent and I don’t know if you even wanted a full explanation, especially since it is pretty much the same method as yours. So, long story short, I don’t see any way of solving it besides using double- or half-angle formulas, but I’m also not a mathematician or a math teacher, just a mediocre student…

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1. Haha, no worries, I totally get what you said. :)

Yeah, I really can’t see any other way than using the double-/half-angle formulae, but I too am just a student, so there might be a better way.

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