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- #1

#### 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

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

what technique should I use here?

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx

- Feb 21, 2013

- 739

Hello Paul,what technique should I use here?

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx

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,

\(\displaystyle |\pi\rangle\)

- Thread starter
- #3

- May 13, 2013

- 386

is this correct?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,

\(\displaystyle |\pi\rangle\)

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.

- Feb 21, 2013

- 739

Hello Paul,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.

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,

\(\displaystyle \pi\rangle\)

- Jan 17, 2013

- 1,667

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 ?

- Jan 17, 2013

- 1,667

- Jan 17, 2013

- 1,667

\(\displaystyle \cos^7 2x = \cos 2x ( 1-\sin^2 2x)^4 \)

The rest is for you .

The rest is for you .

- Thread starter
- #8

- May 13, 2013

- 386

i have a feeling that this correct. is it?

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 ?

=>∫(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

- Jan 17, 2013

- 1,667

Correct !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

- Feb 21, 2013

- 739

Amazing(I never think like that....)\(\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}\)

Sorry Paul for making the problem more work then it need...

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,

\(\displaystyle |\pi\rangle\)

- Jan 17, 2013

- 1,667

\(\displaystyle 1= \sin^2 x + \cos^2 x \)\(\displaystyle \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}\)

How does that work?

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

- Feb 21, 2013

- 739

Hey Zaid,\(\displaystyle 1= \sin^2 x + \cos^2 x \)

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

Thanks for the fast responed...! I see.. My bad that was obvious...

Regards,

\(\displaystyle |\pi\rangle\)

- Admin
- #13

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

Now write the integral as:

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