integration by parts question

**cummings15** · May 5th 2010, 03:27 PM

can someone solve

∫9xcos(3x+1)dx

**ques** · May 5th 2010, 03:30 PM

Originally Posted by cummings15

can someone solve

∫9xcos(3x+1)dx

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

**surjective** · May 5th 2010, 03:34 PM

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:

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

**skeeter** · May 5th 2010, 03:40 PM

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)

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

**ques** · May 5th 2010, 03:43 PM

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

**surjective** · May 5th 2010, 04:05 PM

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.

**ques** · May 5th 2010, 04:07 PM

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.

**cummings15** · May 5th 2010, 04:30 PM

thanks skeeter

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