Help integrating sin^2(x-pi/6)

[K.QMUL]

Hi there everyone,

∫ sin^2(x-pi/6) dx

I have the following integral to solve but am unsure where I should start, I first thought about integrating by parts as I thought you could split it into [Sin(x-pi/6)][Sin(x-pi/6)]. But couldn't seem to figure that out. I was wondering if you could use a trig identity but again am unsure which one.

Any suggestions?

 

[arildno]

Use the double angle formula. This is a very handy formula to reduce the exponent in a trigfunction appearing in your problem.

Integration by parts works nicely as well, if you are careful with your notation.

 

[K.QMUL]

which double angle formula would I use, we still have a Sin^2 to deal with

 

[arildno]

Well, use the double angle formula in which sin^2 appears on its own, of course.

 

[K.QMUL]

Aha, yes, Thanks for the help

 

[K.QMUL]

I have completed the question using the double angle formula, could you tell me if you find any errors, as I am still unsure whether I have done this right or not.

Thanks

 

[Saitama]

I have completed the question using the double angle formula, could you tell me if you find any errors, as I am still unsure whether I have done this right or not.

Thanks

There is a sign error in the end. What is $\displaystyle \int \cos(x) \, dx$?

 

[K.QMUL]

integrating cos(x) is -sin(x) + c, I took that in consideration as it was 1-cos(x), thus using two negatives = positive, I corrected it to x + 1/2 sin(2x- pi/3).

Does everything look good for this integration? Since I checked on Wolfram Mathematica's on-line integration and the answer they got was

http://integrals.wolfram.com/index.jsp?expr=sin^2(x-pi/6)&random=false

 

[K.QMUL]

oh, I realize my mistake, integrating cos(x) = sin(x)

 

[vela]

I have completed the question using the double angle formula, could you tell me if you find any errors, as I am still unsure whether I have done this right or not.

You can always check your answer by differentiating it and seeing if you recover the integrand.

 

[K.QMUL]

So I've completed the question, and checked if I get the original answer by differentiating it. And it seems good. HOWEVER, I have one concern; when you use the double angle formula, can I take 'A' as (x - pi/6) in sin^2(x-pi/6) or would I need to split it somehow. Please clear up my confusion.

 

[vela]

What does 'A' represent?

 

[K.QMUL]

In terms of the question: ^sin2

 

[K.QMUL]

Sorry, in terms of the formula: sin²A = 0.5[1-cos2A]

 

[vela]

In the identity, you can replace 'A' by anything as long as you replace it with the same thing everywhere. Setting A to $(x-\pi/6)$ is perfectly fine.