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

1. Oct 6, 2013

### 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?

2. Oct 6, 2013

### 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.

3. Oct 6, 2013

### K.QMUL

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

4. Oct 6, 2013

### arildno

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

5. Oct 6, 2013

### K.QMUL

Aha, yes, Thanks for the help

6. Oct 6, 2013

### K.QMUL

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

Thanks

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7. Oct 6, 2013

### Saitama

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

8. Oct 6, 2013

### 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

9. Oct 6, 2013

### K.QMUL

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

10. Oct 6, 2013

### vela

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

11. Oct 9, 2013

### K.QMUL

So Ive 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.

12. Oct 9, 2013

### vela

Staff Emeritus
What does 'A' represent?

13. Oct 9, 2013

### K.QMUL

In terms of the question: sin2

14. Oct 9, 2013

### K.QMUL

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

15. Oct 9, 2013

### vela

Staff Emeritus
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.