# Integration help!

#### paulmdrdo

##### Active member
what technique should I use here?

∫dx/1+e^x

∫csc^4x dx/cot^2x

∫cos^7 2x dx

#### Petrus

##### Well-known member
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
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
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

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\displaystyle \cos^7 2x = \cos 2x ( 1-\sin^2 2x)^4$$

The rest is for you .

#### paulmdrdo

##### Active member

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
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
$$\displaystyle \frac{1}{1+e^x}= \frac{1+e^x -e^x}{1+e^x} = 1-\frac{e^x}{1+e^x}$$
Amazing (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
$$\displaystyle \frac{1}{\sin^2 x \, \cos^2 x} =\frac{\sin^2 x + \cos^2 x }{\sin^2 x \, \cos^2 x}$$

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

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

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