MHB Integration Techniques: Solving ∫dx/1+e^x, ∫csc^4x/cot^2x, ∫cos^7 2x

[paulmdrdo1]

what technique should I use here?

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx

 

[Petrus]

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

For the rest I need to think more...

Regards,
$$|\pi\rangle$$

 

[paulmdrdo1]

Re: integration help!

Hello Paul,

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

For the rest I need to think more...

Regards,
$$|\pi\rangle$$

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]

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 $$x - ln(e^x+1) + c.$$ I will solve this as soon as I am home!Regards,
$$\pi\rangle$$

 

[alyafey22]

You missed a sign in the partial fraction .

Here is an another way to do it

$$\frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}$$

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

 

[alyafey22]

$$\Large \frac{\csc^4x}{\cot^2x} = \frac{\frac{1}{\sin^4 x}}{\frac{\cos^2 x}{\sin^2 x}} = \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x} = \sec^2 x + \csc^2 x $$

 

[alyafey22]

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

The rest is for you .

 

[paulmdrdo1]

You missed a sign in the partial fraction .

Here is an another way to do it

$$\frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}$$

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

 

[alyafey22]

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]

$$\frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}$$

Amazing(Clapping)(I never think like that...)
Sorry Paul for making the problem more work then it need...

$$\Large \frac{\csc^4x}{\cot^2x} = \frac{\frac{1}{\sin^4 x}}{\frac{\cos^2 x}{\sin^2 x}} = \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x} = \sec^2 x + \csc^2 x $$

your third and fourth step I did not understand
$$\frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}$$
How does that work? $$|\pi\rangle$$

 

[alyafey22]

$$\frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}$$

How does that work?

$$1= \sin^2 x + \cos^2 x $$

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

 

[Petrus]

$$1= \sin^2 x + \cos^2 x $$

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

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

Regards,
$$|\pi\rangle$$

 

[MarkFL]

Another way to proceed with the second problem is:

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

Now write the integral as:

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