I wanted to know why cos (x/3) intergrates to 3sin (x/3) instead of just sin (x/3)??
Thank you!!
For solving your integral substitute x/3=u, dx/3=du.
$\displaystyle \int\cos(\frac{x}{3}) \ dx$
$\displaystyle \frac{x}{3}=u$
$\displaystyle \frac{dx}{3}=du$
$\displaystyle \int\cos(\frac{x}{3}) \ dx = \int\cos(u) \ (3du)=3\int\cos(u) \ du=3sin(u)+C$
Returning to $\displaystyle x$, by putting $\displaystyle u=\frac{x}{3}$, and we get:
$\displaystyle \int\cos(\frac{x}{3}) \ dx=3sin(\frac{x}{3})+C$
When boromir said "use the chain rule", he was referring to the fact that integration is the reverse of differentiating. Differentiating sin(x/3), using the chain rule, you get cos(x/3) times the derivative of x/3, which is 1/3. That is, the derivative of sin(x/3) would be (1/3)cos(x/3). To get just cos(x/3) as the derivative, you need to multiply by 3 to cancel that 1/3: the derivative of 3sin(x/3) is 3(cos(x/3))(1/3)= cos(x/3).
the chain rule says: (fog)'(x) = f'(g(x))(g'(x)).
so if you are going to integrate something of the form f'(g(x)), you need the g'(x) factor in with the integrand (the thing you're integrating).
in the case of integrating cos(x/3), "f" is the sine funtion, "g" is the function g(x) = x/3. so we need the factor g'(x) = 1/3 under the integral to
make it work. the only way to do this, is to write cos(x/3) = 3[cos(x/3)](1/3), and take the 3 outside the integral.
then [cos(x/3)](1/3) is the desired form f'(g(x))(g'(x)) = (fog)'(x), so integrating THAT will give us (fog)(x), that is sin(x/3).
since the 3 outside the integral is just "along for the ride", we multiply it afterwards. in symbols:
$\displaystyle \int \cos(x/3) dx = \int 3(\cos(x/3))(1/3) dx = 3\int (cos(x/3))(1/3) dx = 3\int (sin(x/3))' dx$
$\displaystyle = 3sin(x/3) + C$