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C3 Integration Help Please. :)

Hello there,

Could someone please tell me how would I integrate 6sin(7y+5) please. I understand for example that 4cos3y when integrated will equal 4/3sin3y but I am unsure on what to do with the bracket.

Thank you. :P
Reply 1
Avatar for ncsjohn02
ncsjohn02
10
dy or dx
Reply 2
Avatar for apo1324
apo1324
OP
11
ncsjohn02
dy or dx


What happens to the (7y+5) part sorry? :s-smilie:
Reply 3
Avatar for Notnek
Notnek
21
apo1324
Hello there,
I understand for example that 4cos3y when integrated will equal 4/3sin3y but I am unsure on what to do with the bracket.

This is correct but why do you understand it?

Have you done integration by substitution? Letting u=3y and following the process will show you why it integrates to (4/3)sin3y.

For your problem, try substituting u=7y+5 and see what happens.

The rules of integration are not always opposite to differentiation so it's not a good idea to just learn integrals without knowing how they're derived.
(edited 13 years ago)
the answer is -6/7cos(7y+5) i think. you just need to multiply your answer by 1/the differential of the bracket
(edited 13 years ago)
Reply 5
Avatar for apo1324
apo1324
OP
11
jimmyreed112
the answer is -6/7cos(7y+5) i think. you just need to multiply your answer by 1/the differential of the bracket


Oh ok, but the answer is -6sin(7y+5)... I don't understand where the minus sign came from.
Reply 6
Avatar for Scott3142
Scott3142
jimmyreed112 is correct for this example, although this will not work for polynomials above degree 1 in the bracket. I'm not entirely sure that if you had a polynomial above degree 1 that it can be integrated with elementary functions at all...
Reply 7
Avatar for apo1324
apo1324
OP
11
notnek
This is correct but why do you understand it?

Have you done integration by substitution? Letting u=3y and following the process will show you why it integrates to (4/3)sin3y.

For your problem, try substituting u=7y+5 and see what happens.

The rules of integration are not always opposite to differentiation so it's not a good idea to just learn integrals without knowing how they're derived.


No, I haven't done integration by substitution. :s-smilie: Yeah, basically my teacher gave us a list of all the integrals and expects us to learn them.
Reply 8
Avatar for Scott3142
Scott3142
The minus sign comes from the fact that the integral of sin is -cos, that's all.
the integral of sinx is -cosx...its just a case of learning the rules of integrating and differentiating trigonometric functions
Reply 10
Avatar for apo1324
apo1324
OP
11
For this example... -2cos(10-x):-

Would it be -2sin(10-x) + c as although 10-x differentiated = -1 cos becomes a negative again when integrating sin?? Thanks for the help everyone, appreciate it btw. :smile: so -2/1sin(10-x) + c.
(edited 13 years ago)
the integral of -2cos(10-x) is 2sin(10-x) + c and yes youre right
Reply 12
Avatar for apo1324
apo1324
OP
11
jimmyreed112
the integral of -2cos(10-x) is 2sin(10-x) + c and yes youre right


Thanks. :smile:
apo1324
Thanks. :smile:

anytime:smile:
Reply 14
Avatar for nuodai
nuodai
17
If you want to do it another (and safer) way, use the fact that sine integrates to some multiple of cosine, and cosine integrates to some multiple of sine (when what's in the brackets is linear, but in C3 it will always be linear).

So you can let ∫sin⁡(ax+b) dx=Kcos⁡(ax+b)+C\displaystyle \int \sin(ax+b)\, dx = K\cos(ax+b) + C. Then differentiate the RHS and compare it to sin⁡(ax+b)\sin (ax+b) to find the value of KK.

So for example let's say we have to integrate cos⁡(3−11x)\cos (3-11x). Then we can say:
∫cos⁡(3−11x) dx=Ksin⁡(3−11x)+C\displaystyle \int \cos(3-11x)\, dx = K\sin (3-11x) + C

Differentiating the both sides gives
cos⁡(3−11x)=−11Kcos⁡(3−11x)\cos (3-11x) = -11K\cos(3-11x)

...and so we must have −11K=1-11K=1, and hence K=−111K=-\dfrac{1}{11}, so our final answer is:

∫cos⁡(3−11x) dx=−111sin⁡(3−11x)+C\boxed{ \displaystyle \int \cos(3-11x)\, dx = -\dfrac{1}{11}\sin(3-11x) + C }

Hope this helps.

[EDIT: Thanks ghostwalker for the correction]
(edited 13 years ago)

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