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Help with trigonometric integration question

I was thinking to do integration by parts, but the mark scheme says otherwise and I don't understand the mark scheme. Please explain, thanks!

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Reply 1
Avatar for joostan
joostan
13
Original post by DoeADeer
I was thinking to do integration by parts, but the mark scheme says otherwise and I don't understand the mark scheme. Please explain, thanks!

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Is that sin⁡(x)\sin(x) before the bracket? In which case I'd recommend expanding the brackets :smile:

EDIT: As to what the markscheme's done it's done the reverse chain rule.
Reply 2
Avatar for DoeADeer
DoeADeer
OP
9
Original post by joostan
Is that sin⁡(x)\sin(x) before the bracket? In which case I'd recommend expanding the brackets :smile:

EDIT: As to what the markscheme's done it's done the reverse chain rule.


Yeah, sinx. How'd I integrate sinxcosx though?! I am tearing my hair out at this simple question aghh.

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Reply 3
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Another
21
sinxcosx is the same as 1/2sin2x (double angle formulae.... sort of.)


EDIT: Looking at the image of the markscheme... lol what on earth did they do there???
(edited 10 years ago)
Reply 4
Avatar for joostan
joostan
13
Original post by DoeADeer
Yeah, sinx. How'd I integrate sinxcosx though?! I am tearing my hair out at this simple question aghh.

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What if I told you sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)
Would that be enough of a hint? :wink:
Reply 5
Avatar for DoeADeer
DoeADeer
OP
9
Original post by Another
sinxcosx is the same as 1/2sin2x (double angle formulae.... sort of.)


Original post by joostan
What if I told you sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)
Would that be enough of a hint? :wink:


I still haven't got it.
Could you explain how I've gone wrong please??

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Reply 6
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Another
21
Original post by DoeADeer
I still haven't got it.
Could you explain how I've gone wrong please??

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integral of sinx is -cosx, negative not positive

You don't need to change cos2x at all btw, (although it should still give you the same answer. Not wrong, just long.) Just substitute pi and 0
Original post by joostan
What if I told you sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)
Would that be enough of a hint? :wink:


Using your method, I get 1/4 cos2x - cosx + c, is this what you got? However, the mark scheme has a different result.
Reply 8
Avatar for joostan
joostan
13
Original post by Lackadaisical
Using your method, I get 1/4 cos2x - cosx + c, is this what you got? However, the mark scheme has a different result.


The markscheme uses the reverse chain rule. I've not actually checked the results, but what you should find is that the solutions are the same, only the + c is different.
For a definite integral, this won't make any difference :smile:

EDIT: Yep - the constant is different but the forms are equivalent. :smile:
(edited 10 years ago)
Reply 9
Avatar for Another
Another
21
Original post by Lackadaisical
Using your method, I get 1/4 cos2x - cosx + c, is this what you got? However, the mark scheme has a different result.


That's -1/4cos2x - cosx. The results are the same, I've checked it :biggrin:
Original post by Another
That's -1/4cos2x - cosx. The results are the same, I've checked it :biggrin:


Yes, of course, ok :smile:

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