# Thread: Integral of tan^3(x) dx where did i go wrong?

1. ## Integral of tan^3(x) dx where did i go wrong?

Hi

I am trying to solve the integral of tan^3(x) dx but i seem to have made a slight error with the answer.

This is the calculations i made, i can't see where my logic failed me:

The answer seems to be natural log of cos x instead of sec x. But i am not quite getting how ?

Hope some one can spot my error.

2. ## Re: Integral of tan^3(x) dx where did i go wrong?

$\begin{gathered} \int {{{\tan }^3}(x)dx} \hfill \\ \int {{{\tan }^2}} (x)\tan (x)dx \hfill \\ \int {({{\sec }^2}(x) - 1)\tan (x)dx} \hfill \\ \int {(\tan (x){{\sec }^2}(x) - \tan (x))dx} \hfill \\ \frac{1}{2}{\tan ^2}(x) + \log (\cos (x)) \hfill \\ \end{gathered}$

3. ## Re: Integral of tan^3(x) dx where did i go wrong?

Originally Posted by thefollower
Hope some one can spot my error.
There isn't one, I don't think.
$$\ln|\sec x|=\ln \left|\frac1{\cos x}\right|=\ln|\cos x|^{-1}=-\ln|\cos x|$$
Plato has a sign error in his integral of $\int \tan x \mathrm dx = \ln |\cos x|$

4. ## Re: Integral of tan^3(x) dx where did i go wrong?

Originally Posted by Archie
There isn't one, I don't think.
$$\ln|\sec x|=\ln \left|\frac1{\cos x}\right|=\ln|\cos x|^{-1}=-\ln|\cos x|$$
Plato has a sign error in his integral of $\int \tan x \mathrm dx = \ln |\cos x|$
Oh really ? I got marked wrong which is kinda annoying. I am also confused how Plato got 1/2 tan^2(x) not 1/2sec^2(x) for the first term.

5. ## Re: Integral of tan^3(x) dx where did i go wrong?

Originally Posted by thefollower
Oh really ? I got marked wrong which is kinda annoying. I am also confused how Plato got 1/2 tan^2(x) not 1/2sec^2(x) for the first term.
You do this and post the result:
The derivative of $\dfrac{1}{2}\tan^2(x)$ is:

6. ## Re: Integral of tan^3(x) dx where did i go wrong?

Originally Posted by thefollower
I am also confused how Plato got 1/2 tan^2(x) not 1/2sec^2(x) for the first term.
$$\frac{\mathrm d}{\mathrm dx}\tan x = \sec^2 x$$
So, by writing $t=\tan x$ we get $\mathrm dt = \sec^2 x\, \mathrm d x$ and perform the substitution.

7. ## Re: Integral of tan^3(x) dx where did i go wrong?

Originally Posted by thefollower
Oh really ? I got marked wrong ...
Other than that vertical bar next to the constant of integration (which
looks like a typo), the derivative of your anti-derivative simplifies
to the original expression that you were
taking the integral of. The instructor should
go back and give you credit for the problem.