1. cal II intergral problem

am running into trouble intergrating
the integral tan^5(x) dx

i know i need to break it up in terms of sec(x)^2 and tan(x)sec(x) so that i can use u subsitution

also would you guys give me any tips on how to do well in cal II am taking a summer course right now
and i need all the tips that i can get thank you very much

2. Originally Posted by momo55
am running into trouble intergrating
the integral tan^5(x) dx

i know i need to break it up in terms of sec(x)^2 and tan(x)sec(x) so that i can use u subsitution

also would you guys give me any tips on how to do well in cal II am taking a summer course right now
and i need all the tips that i can get thank you very much
$\displaystyle \tan^5{x} = \tan^3{x}\tan^2{x}$

$\displaystyle = \tan^3{x}(\sec^2{x} - 1)$

$\displaystyle = \tan^3{x}\sec^2{x} - \tan^3{x}$

$\displaystyle = \tan^3{x}\sec^2{x} - \tan{x}(\sec^2{x} - 1)$

$\displaystyle = \tan^3{x}\sec^2{x} - \tan{x}\sec^2{x} + \tan{x}$

$\displaystyle = \sec^2{x}(\tan^3{x} - \tan{x}) + \tan{x}$.

So $\displaystyle \int{\tan^5{x}\,dx} = \int{\sec^2{x}(\tan^3{x} - \tan{x}) + \tan{x}\,dx}$

$\displaystyle = \int{\sec^2{x}(\tan^3{x} - \tan{x})\,dx} - \int{\frac{-\sin{x}}{\phantom{-}\cos{x}}\,dx}$.

Now let $\displaystyle u = \tan{x}$ so that $\displaystyle du = \sec^2{x}\,dx$ and let $\displaystyle v = \cos{x}$ so that $\displaystyle dv = -\sin{x}\,dx$, and the integrals become

$\displaystyle \int{u^3 - u\,du} - \int{\frac{1}{v}\,dv}$

$\displaystyle = \frac{1}{4}u^4 - \frac{1}{2}u^2 - \ln{|v|} + C$

$\displaystyle = \frac{1}{4}\tan^4{x} - \frac{1}{2}\tan^2{x} - \ln{|\cos{x}|} +C$.

sec(x)^4/4 - tan(x)^2/2 - ln|sec(x)|

i just don't know how to get there

4. IF you use the identity with tangent and secant, that was used to integrate your problem,
you will be able to show that the two answers are within a constant (C), of being the same.

However, one of these seem off, with the constant in front of the logarithm.

$\displaystyle \ln |\sec x|=-\ln |\cos x|$

5. Originally Posted by momo55
sec(x)^4/4 - tan(x)^2/2 - ln|sec(x)|

i just don't know how to get there
It should be a + in front of the logarithm there...

$\displaystyle -\ln{|\cos{x}|} = -\ln{\left|\frac{1}{\sec{x}}\right|}$

$\displaystyle = -\ln{|\sec{x}|^{-1}}$

$\displaystyle = \ln{|\sec{x}|}$.

6. yea i want to know the steps to get there

7. Originally Posted by momo55
yea i want to know the steps to get there