1. Integration by Substitution

$\displaystyle \int 5\cos(5x) dx$

$\displaystyle g'(x) = 5, f(g(x)) = cos5x$

$\displaystyle \int (cos5x)(5) dx$

$\displaystyle (1/5) \sin5x + C$

Those are my steps above, but the book says the answer is sin5x. It seems like it got (1/5)*sin5x*5 after integrating and then canceled the 5 and 1/5, but I thought that $\displaystyle \int f(g(x))g'(x) = F(g(x)) + C$, which to me means that you drop the g'(x) and find the antiderivative of f(g(x)). My other thought is that I just find the antiderivative of f(u), which would be cosu in this case, which yields sinu + C. Am I correct?

2. Originally Posted by tom ato
$\displaystyle \int 5\cos(5x) dx$

$\displaystyle g'(x) = 5, f(g(x)) = cos5x$

$\displaystyle \int (cos5x)(5) dx$

$\displaystyle (1/5) \sin5x + C$

Those are my steps above, but the book says the answer is sin5x. It seems like it got (1/5)*sin5x*5 after integrating and then canceled the 5 and 1/5, but I thought that $\displaystyle \int f(g(x))g'(x) = F(g(x)) + C$, which to me means that you drop the g'(x) and find the antiderivative of f(g(x)). My other thought is that I just find the antiderivative of f(u), which would be cosu in this case, which yields sinu + C. Am I correct?
$\displaystyle \int (cos5x)(5) dx$

=$\displaystyle (1/5) \sin5x(5) + C$

=$\displaystyle sin 5x+C$

3. You can take constants outside the integration sign.

$\displaystyle 5 \int \cos(5x) dx = 5 \bigg{(}\sin(5x) \cdot \frac{1}{5} \bigg{)} + C$

(note that you don't have to multiply the constant by 5 because the result will still be a constant.)

Then cancel the 5 and 1/5

4. Just noticed the mistake in your answer, you only integrate f(x) which is cos(x)

Then you 'sub' in x -> g(x). So you get... sin(x) -> sin(5x)

I think the proper way to write it is...

$\displaystyle \int f(g(t))g'(t) dt = \int f(x) dx = F(x) + c$

where x = g(t)