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Integration help!

paulmdrdo

Active member
May 13, 2013
386
what technique should I use here?

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx
 

Petrus

Well-known member
Feb 21, 2013
739
Re: integration help!

what technique should I use here?

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx
Hello Paul,

For the first one I asume you mean
\(\displaystyle \int\frac{dx}{1+e^x}\)
start with multiply both side with \(\displaystyle e^x\) so we got:
\(\displaystyle \int\frac{e^x}{e^x(1+e^x)}dx\)
subsitute \(\displaystyle u=e^x+1 <=> du=e^xdx\)
so our integrate become
\(\displaystyle \int \frac{du}{(u-1)u}\)
I leave the partial fraction to you:)

For the rest I need to think more...

Regards,
\(\displaystyle |\pi\rangle\)
 

paulmdrdo

Active member
May 13, 2013
386
Re: integration help!

Hello Paul,

For the first one I asume you mean
\(\displaystyle \int\frac{dx}{1+e^x}\)
start with multiply both side with \(\displaystyle e^x\) so we got:
\(\displaystyle \int\frac{e^x}{e^x(1+e^x)}dx\)
subsitute \(\displaystyle u=e^x+1 <=> du=e^xdx\)
so our integrate become
\(\displaystyle \int \frac{du}{(u-1)u}\)
I leave the partial fraction to you:)

For the rest I need to think more...

Regards,
\(\displaystyle |\pi\rangle\)
is this correct?

1/(u*(u-1)) = 1/(u-1) + 1/u
ln(u-1) + ln(u) + c
= ln(e^x) + ln(e^x+1) + c
= x + ln(e^x+1) + c.
 

Petrus

Well-known member
Feb 21, 2013
739
Re: integration help!

is this correct?

1/(u*(u-1)) = 1/(u-1) + 1/u
ln(u-1) + ln(u) + c
= ln(e^x) + ln(e^x+1) + c
= x + ln(e^x+1) + c.
Hello Paul,
Right now I am not home but the correct answer (with wolframalpha) is \(\displaystyle x - ln(e^x+1) + c.\) I will solve this as soon as I am home!


Regards,
\(\displaystyle \pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You missed a sign in the partial fraction .

Here is an another way to do it

\(\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}\)

Now you realize that the numerator is the derivative of the denominator , can you finish now ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \Large \frac{\csc^4x}{\cot^2x} = \frac{\frac{1}{\sin^4 x}}{\frac{\cos^2 x}{\sin^2 x}} = \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x} = \sec^2 x + \csc^2 x \)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \cos^7 2x = \cos 2x ( 1-\sin^2 2x)^4 \)

The rest is for you .
 

paulmdrdo

Active member
May 13, 2013
386
You missed a sign in the partial fraction .

Here is an another way to do it

\(\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}\)

Now you realize that the numerator is the derivative of the denominator , can you finish now ?
i have a feeling that this correct. is it?
=>∫(e^x+1)/e^x+1) dx - ∫e^x dx /(e^x+1)

=>∫dx - ∫d(e^x+1)/e^x+1

=>x - ln(e^x+1) + c
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
i have a feeling that this correct. is it?
=>∫(e^x+1)/e^x+1) dx - ∫e^x dx /(e^x+1)

=>∫dx - ∫d(e^x+1)/e^x+1

=>x - ln(e^x+1) + c
Correct !
 

Petrus

Well-known member
Feb 21, 2013
739
\(\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}\)
Amazing(Clapping)(I never think like that....)
Sorry Paul for making the problem more work then it need...



\(\displaystyle \Large \frac{\csc^4x}{\cot^2x} = \frac{\frac{1}{\sin^4 x}}{\frac{\cos^2 x}{\sin^2 x}} = \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x} = \sec^2 x + \csc^2 x \)
your third and fourth step I did not understand
\(\displaystyle \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}\)
How does that work?

Best regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}\)

How does that work?
\(\displaystyle 1= \sin^2 x + \cos^2 x \)

I substituted instead of \(\displaystyle 1\) in the numerator the term \(\displaystyle \sin^2 x + \cos^2 x \).
 

Petrus

Well-known member
Feb 21, 2013
739
\(\displaystyle 1= \sin^2 x + \cos^2 x \)

I substituted instead of \(\displaystyle 1\) in the numerator the term \(\displaystyle \sin^2 x + \cos^2 x \).
Hey Zaid,
Thanks for the fast responed...! I see.. My bad that was obvious...

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Another way to proceed with the second problem is:

\(\displaystyle \frac{\csc^4(x)}{\cot^2(x)}=\frac{\tan^2(x)}{\sin^4(x)}=\frac{1}{\left(\sin(x)\cos(x) \right)^2}=4\csc^2(2x)\)

Now write the integral as:

\(\displaystyle -2\int-\csc^2(2x)\,2\,dx\)