# Integration by Substitution

• Jun 12th 2009, 07:21 AM
tom ato
Integration by Substitution
$
\int 5\cos(5x) dx
$

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

$
\int (cos5x)(5) dx
$

$
(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 $\int f(g(x))g'(x) = F(g(x)) + C$, 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?
• Jun 12th 2009, 07:27 AM
alexmahone
Quote:

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

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

$
\int (cos5x)(5) dx
$

$
(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 $\int f(g(x))g'(x) = F(g(x)) + C$, 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?

$\int (cos5x)(5) dx$

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

= $sin 5x+C$
• Jun 12th 2009, 07:27 AM
You can take constants outside the integration sign.

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

(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
• Jun 12th 2009, 07:32 AM
$\int f(g(t))g'(t) dt = \int f(x) dx = F(x) + c$