[itex]\int tan^{5}x dx[/itex] , Two Methods, are both correct?

[Lebombo]

Homework Statement

Are both methods and final results correct?$\int tan^{5}xdx$Method 1:

$\int tan^{3}xtan^{2}xdx$

$\int tanx^{3} (sec^{2}x - 1) dx$

$\int tan^{3}xsec^{2}x dx - \int tanxtan^{2}x dx$

$\int tan^{3}xsec^{2}x dx - \int tanxsec^{2}x dx + \int tanx$

$\int u^{3}du - \int u du - \int \frac{1}{u} du$

$\frac{1}{4}tan^{4}x - \frac{1}{2}tanx - ln|cosx| + C$
Method 2:

$\int tan^{5}x dx$

$\int tan^{4}tanx$

$\int (sec^{2}x - 1)^{2}tanxdx$

$\int (sec^{4}x - 2sec^{2}x + 1)tanx dx$

$\int (u^{2} - 2u +1) du$

$\frac{1}{3}tan^{3}x - tan^{2}x + tanx + C$

 

[Yosty22]

Although Method 2 is much easier and a lot more straightforward, both methods look correct, despite having different answers.

 

[Lebombo]

Thanks Yosty.

 

[arildno]

I don't understand what you are doing.
Tip:
Make absolutely clear WHAT substitution you are using!
Here is how I would do it:
$$u=\tan(x)\to{du}=\sec^{2}(x)dx=(u^{2}+1)dx \to{dx}=\frac{du}{u^{2}+1}$$
Thus, you are to evaluate the integral:
$$\int\frac{u^{5}}{u^{2}+1}du= \int\frac{u^{3}(u^{2}+1-1)}{u^{2}+1}du =\int{u^{3}}du-\int\frac{u^{3}}{u^{2}+1}du$$
The last integral can be written as:
$$-\int\frac{u^{3}}{u^{2}+1}du=-\int{u}du+\int\frac{u}{u^{2}+1}du$$
Thus, collecting together, you'll have:
$$\frac{1}{4}\tan^{4}(x)-\frac{1}{2}\tan^{2}(x)+\frac{1}{2}\ln(\tan^{2}(x)+1)$$
where the latter term can be simplified to $-\ln|\cos(x)|$

 

[arildno]

Thus, NEITHER of your expressions are correct, although the first comes closer.

 

[D H]

Method 2 is also incorrect. Your u-substitution, $u=\sec x$ lost something in the translation. With this, you should get $du = \sec x \tan x\, dx$, but you apparently used $du = \tan x \, dx$.

 

[Lebombo]

Thank you Arildno,


Method 1:

$\int tan^{3}xtan^{2}xdx$

$\int tanx^{3} (sec^{2}x - 1) dx$

$\int tan^{3}xsec^{2}x dx - \int tanxtan^{2}x dx$

$\int tan^{3}xsec^{2}x dx - \int tanxsec^{2}x dx + \int tanx dx$


I did three separate substitutions.

For $\int tan^{3}xsec^{2}x dx$

I let u = tanx, and $du = sec^{2}x dx$

For $\int tanxsec^{2}x dx$

I let u = tanx, and $du = sec^{2}x dx$

For $\int tanx dx$,

I solved by writing as $\int \frac{sinx}{cosx} dx$

and then letting u = cosx, so $-du = sinx dx$

This produces:

$\int u^{3} du - \int u du - \int \frac{1}{u} du$

 

[arildno]

Well, int udu goes to 1/2u^2, doesn't it?, rather than your 1/2u?