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Core Maths 3: How do you integrate this?

Hi there

I have the question:

∫sinx3dx\displaystyle \int sin \frac{x}{3} dx

I know that sin x integrated is - cos x,

so is the answer to my question, before working out the exact values with the limits, this;

−cosx3+c\displaystyle - cos \frac{x}{3} + c

Many thanks
(edited 12 years ago)
Reply 1
Avatar for gridlock
gridlock
8
The integral of sin(ax+b) is -(1/a)cos(ax+b).
That means in this situation a is 1/3.
Reply 2
Avatar for jackie11
jackie11
OP
4
ok so in my case its -3 cos x ??
Reply 3
Avatar for gridlock
gridlock
8
The x value must remain the same as 'a' remains in front it.
Reply 4
Avatar for jackie11
jackie11
OP
4
so its -3 cos 1/3x which is - cos x??
Reply 5
Avatar for gridlock
gridlock
8
Yes and no. Cos(ax) is one value and 'a' cannot be changed by the value cos is multiplied by. The answer is -3cos(x/3). There's also a '+ c' but you already knew that. :P
(edited 12 years ago)
Reply 6
Avatar for jackie11
jackie11
OP
4
ok thank you very much :smile:
Reply 7
Avatar for gridlock
gridlock
8
Original post by jackie11
ok thank you very much :smile:


No problem; thanks for the rep. :smile:
Reply 8
Avatar for Pin
Pin
Hey
I would suggest using the substitution method.
x3=u \frac{x}{3} = u therefore dudx=13\frac{du}{dx} = \frac{1}{3} so dx = 3 du

The original equation is ∫sinx3dx \displaystyle \int sin \frac{x}{3}dx if you put in all the substituions you should get
∫sinu3du \displaystyle \int sin u 3du
which simplifies to 3∫sinudu \displaystyle 3 \int sin u du

This should be fine to solve and then replace u with x/3 to get the final answer.

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