# integral

• June 11th 2013, 01:06 AM
kastamonu
integral
integrate dx/tanxcos^2x

sinx=u and du=cosxdx and dxcosx/sinxcos^2x then du/u(1-u)^2
1/u+1/1-u+1/1+u
a/u + b/1-u + c/1+u
I found c=-1/2, a=1 and b=1/2

ln|u|+1/2ln|1-u|-1/2ln|1+u|

ln|sinx|+1/2ln|1-sinx|-1/2ln|1+sinx|
ln|sinx|+ln|1-sinx|/1/2ln|1+sinx|

and I got stuck here.How can I solve that?
• June 11th 2013, 02:00 AM
ibdutt
Re: integral
• June 11th 2013, 05:43 AM
kastamonu
Re: integral
Iwant to solve it in my way.
• June 11th 2013, 06:22 AM
Plato
Re: integral
Quote:

Originally Posted by kastamonu
Iwant to solve it in my way.

Then why did you ask for help?

Also, why don't you learn to post in a readable format.

Here is what you actually posted $\int {\frac{{dx}}{{\tan (x)}}{{\cos }^2}(x)}$.

But it appears that is not what you mean, If so, learn to use grouping symbols.

Now note that $\int {\frac{{dx}}{{\tan (x)}\cos^2(x)} =\int {\frac{{\sec^2(x)dx}}{{\tan (x)}}$

If you do not recognize at once that as a logarithm, then you are crippled by u-substitutions.
• June 11th 2013, 06:40 AM
HallsofIvy
Re: integral
For example, "b/1- u" would, strictly, be read "b- u" but I am sure you mean b/(1- u).

Quote:

Originally Posted by kastamonu
integrate dx/tanxcos^2x

sinx=u and du=cosxdx and dxcosx/sinxcos^2x then du/u(1-u)^2
1/u+1/1-u+1/1+u
a/u + b/1-u + c/1+u
I found c=-1/2, a=1 and b=1/2

ln|u|+1/2ln|1-u|-1/2ln|1+u|

ln|sinx|+1/2ln|1-sinx|-1/2ln|1+sinx|
ln|sinx|+ln|1-sinx|/1/2ln|1+sinx|

A typo. You mean ln|sin(x)|+ (1/2)ln(|1- sin(x)|)- (1/2)ln(|1+ sin(x)|)

Quote:

and I got stuck here.How can I solve that?
What do you mean by "solve" it? You were asked to find an integral and you have. I presume you mean "simplify it" or write it in a simpler form. To do that use the laws of logarithms:
$cln(a)= ln(a^c)$ so $(1/2)ln(|1- sin(x)|)= ln(|1- sin(x)|^{1/2})= ln(\sqrt{1- sin(x)}$
and $(1/2)ln(|1+ sin(x)|)= ln(|1+ sin(x)|^{1/2}= ln(\sqrt{1+ sin(x)})$

$ln(a)- ln(b)= ln(\frac{a}{b})$ so
$(1/2)ln(|1- sin(x)|)- (1/2)ln(|1+ sin(x)|)= ln(\sqrt{1- sin(x)})- ln(\sqrt{1+ sin(x)})= ln\left(\frac{\sqrt{1- sin(x)}}{\sqrt{1+ sin(x)}}\right)= ln\left(\sqrt{\frac{1- sin(x)}{1+ sin(x)}}\right)$

$ln(a)+ ln(b)= ln(ab)$ so
$ln|sin(x)|+ (1/2)ln(1- sin(x))- (1/2)ln(1+ sin(x))= ln|sin(x)|ln\left(\sqrt{\frac{1- sin(x)}{1+ sin(x)}}\right)= ln\left|sin(x)\sqrt{\frac{1- sin(x)}{1+ sin(x)}}\right|$
• June 12th 2013, 02:46 AM
kastamonu
Re: integral
I was stuck. By this reason I asked for your help. I don't like memorizing formulas so I wanted to solve it in that way. I made a mistake in "1/1-u".I found the answer.Many Thanks.
• June 12th 2013, 02:47 AM
kastamonu
Re: integral
Quote:

Originally Posted by Plato
Then why did you ask for help?

Also, why don't you learn to post in a readable format.

Here is what you actually posted $\int {\frac{{dx}}{{\tan (x)}}{{\cos }^2}(x)}$.

But it appears that is not what you mean, If so, learn to use grouping symbols.

Now note that $\int {\frac{{dx}}{{\tan (x)}\cos^2(x)} =\int {\frac{{\sec^2(x)dx}}{{\tan (x)}}$

If you do not recognize at once that as a logarithm, then you are crippled by u-substitutions.

I agree with you that it is pointless to remember too many formulas, as they simply get mixed up and end up confusing people. However, $\displaystyle \frac{d}{dx} \left[ \tan{(x)} \right] = \sec^2{(x)}$ is a standard derivative, so is one that I and you should memorise.