# $\int tan^{5}x dx$ , Two Methods, are both correct?

• Lebombo
In summary: 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 + CAlthough Method 2 is much easier and a lot more straightforward, both methods look correct, despite having different answers.Thanks Yosty.In summary, both methods are correct, but Method

## 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:

$\int tan^{5}x dx$

$\int tan^{4}tanx$

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

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

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

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

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

1 person
Thanks Yosty.

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 $-\ln|\cos(x)|$

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

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##.

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:

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

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

## 1. What is the formula for finding the integral of tan5(x)?

The formula for finding the integral of tan5(x) is: $\int tan^{5}x dx = \frac{tan^{6}x}{6} - \frac{tan^{4}x}{4} + C$

## 2. Can you use both methods to solve the integral of tan5(x)?

Yes, both methods (integration by parts and substitution) can be used to solve the integral of tan5(x). However, the choice of method may depend on the complexity of the integral and personal preference.

## 3. What is the first step in solving the integral of tan5(x) using integration by parts?

The first step in solving the integral of tan5(x) using integration by parts is to identify the parts of the integrand that can be differentiated and integrated separately. In this case, we can set u = tan4(x) and dv = tan(x) dx.

## 4. How do you know if your answer is correct when using substitution to solve the integral of tan5(x)?

To check if your answer is correct when using substitution to solve the integral of tan5(x), you can differentiate your answer and see if it matches the original integrand. You can also use a calculator or online integral calculator to verify your answer.

## 5. Can you use any other methods to solve the integral of tan5(x)?

Yes, there are other methods that can be used to solve the integral of tan5(x) such as trigonometric identities, partial fractions, and series expansion. However, these methods may not always be the most efficient or practical for this particular integral.