integration by parts

makuna · #1 $Old$ June 29th, 2009, 12:46 PM

hey guys i need some help with this problem. i did it but a got a really bad answer.

∫tanx dx= ∫sinx/cosx dx
u=1/cosx dv=sinx dx
du=tanx secx dx v=-cosx

∫tanx dx= uv-∫vdu
∫tanx dx=-1+∫tanx dx
0=-1

if u guys can guide me will be awesomew

chengbin · #2 $Old$ June 29th, 2009, 01:07 PM

This does not require integration by part. It is an integration by substitution problem. Here is a shorter way of doing it, assuming you know the formula $\int \frac {f'(x)} {f(x)}dx = \ln |f(x)|+c$

$\int \tan xdx$

$\int \frac {\sin x}{\cos x}dx$

$-\ln|\cos x|+c$

If you want to do integration by substitution, let u= $\cos x$

makuna · #3 $Old$ June 29th, 2009, 01:10 PM

thanks for the reply.
the thing is that when i do that i get.

-ln|cos x|+c=-1+ln|cos x|+c

chengbin · #4 $Old$ June 29th, 2009, 01:12 PM

Quote:

Originally Posted by makuna $View Post$

thanks for the reply.
the thing is that when i do that i get.

-ln|cos x|+c=-1+ln|cos x|+c

Show us what you did, especially how you got the -1.

makuna · #5 $Old$ June 29th, 2009, 01:16 PM

sure
∫tanx dx= ∫sinx/cosx dx
u=1/cosx dv=sinx dx
du=tanx secx dx v=-cosx

∫tanx dx= uv-∫vdu
∫tanx dx=1/cosx(-cosx)- ∫-cosx(tanx . secx)
∫tanx dx=-1+∫tanx dx
∫tanx dx -∫tanx dx=-1
0=-1

Bruno J. · #6 $Old$ June 29th, 2009, 01:31 PM

I believe you're just confused about the meaning of the constant of integration. The constant of integration is not really a constant - it satisfies weird identities like c+1=c. The indefinite integral of $f(x)$ is an infinite class of functions whose derivatives are all $f(x)$ . It's not really a function of $x$ itself. When you say

∫tanx dx -∫tanx dx=-1

this is not actually meaningless - what it means is that the difference of two functions whose derivative is $\tan x$ is a constant function. The left-hand side is not really zero, it's an infinite class of functions - in this case, the class of functions with zero derivative, i.e. the class of constant functions. If you had

∫tanx dx -∫tanx dx = x

then you'd be in trouble, because the left-hand side has zero derivative, and the right-hand side has a derivative of 1. But

∫tanx dx -∫tanx dx=-1

is not a contradiction.

To look at it in another way, think of the indefinite integral $\int 0 \: dx$ ; in fact $\int 0 \: dx = C$ is the class of constant functions.

chengbin · #7 $Old$ June 29th, 2009, 01:43 PM

The correct way to do this problem is like this.

$\tan xdx$

$\frac {\sin x}{\cos x}dx$

$\text {Let} u=\cos x$

$du=-\sin xdx$

$\int -\frac {du}{u}$

$-\ln |u|+c$

$-\ln |\cos x|+c$

makuna · #8 $Old$ June 29th, 2009, 01:54 PM

ok, well when i integrate ∫tanx dx with Π/6,Π/4. i get the same answer.
0=-1

HallsofIvy · #9 $Old$ June 29th, 2009, 02:55 PM

Are you paying any attention at all to the responses? The very first response suggested NOT using integration by parts and you said "The thing is when I do that I get" and just gave the same answer as before!

If you insist upon using integration by parts:
Try u= sin x, dv= sec x dx. Then du= cos x dx and v=
Let u= sec x so du= sec x tan x dx , dv= sin x dx so v= -cos x.
$\int tan x dx= -sec x cos x+ \int cos x sec x tan x dx$
$\int tan x dx= -1+ \int tan x dx+ C$ which reduces to the true statement 0= -1+ C. That doesn't give the integral but that's because integration by parts is not a good method for this.