# Math Help - integration by parts question

1. ## integration by parts question

can someone solve

∫9xcos(3x+1)dx

2. Originally Posted by cummings15
can someone solve

∫9xcos(3x+1)dx
= 9x∫cos(3x + 1)dx - ∫d/dx (9x) ∫cos(3x + 1)dx dx

3. ## integration

Hello,

Define the functions $f(x)=9x$ and $g(x)=\cos(3x-1)$.

Choose which function you would like to integrate and which one you would like to diferentiate. Then simply apply the formula:

$\int f(x)g(x)= F(x)g(x)-\int F(x)g'(x)$

4. Originally Posted by cummings15
can someone solve

∫9xcos(3x+1)dx

tabular integration ...

sign ....... u ............ dv .....

(+) .........x ........... 9cos(3x+1)

(-) .........1 ........... 3sin(3x+1)

..............0............-cos(3x+1)

$\int 9x\cos(3x+1) \, dx = 3x\sin(3x+1) + \cos(3x+1) + C
$

5. Originally Posted by surjective
Hello,

Define the functions $f(x)=9x$ and $g(x)=\cos(3x-1)$.

Choose which function you would like to integrate and which one you would like to diferentiate. Then simply apply the formula:

$\int f(x)g(x)= F(x)g(x)-\int F(x)g'(x)$
It would not be wise to choose 9x as the function to integrate.

The order in which the first(to differentiate) and second(to integrate) functions should be taken is

Inverse Trigonometric
Logarithmic
Algebraic
Trigonometric
Exponential

6. ## integration

Hello,

The functions I defined were not meant to be fixed. You could just swop them around. I just wanted the basic idea to be understood, i.e. that you may choose either one of the function to integrate or differentiate, ofcourse selecting wisely.

7. Originally Posted by surjective
Hello,

The functions I defined were not meant to be fixed. You could just swop them around. I just wanted the basic idea to be understood, i.e. that you may choose either one of the function to integrate or differentiate, ofcourse selecting wisely.

8. thanks skeeter