cal II intergral problem

**momo55** · June 5th, 2010, 05:50 PM

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

**Prove It** · June 5th, 2010, 05:58 PM

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

$\tan^5{x} = \tan^3{x}\tan^2{x}$

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

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

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

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

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

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

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

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

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

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

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

**momo55** · June 5th, 2010, 06:43 PM

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

i just don't know how to get there

**matheagle** · June 5th, 2010, 06:57 PM

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.

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

**Prove It** · June 5th, 2010, 07:01 PM

Originally Posted by momo55

the answer is
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...

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

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

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

**momo55** · June 5th, 2010, 07:07 PM

yea i want to know the steps to get there

**Prove It** · June 5th, 2010, 07:14 PM

Originally Posted by momo55

yea i want to know the steps to get there

Read my posts...

**momo55** · June 5th, 2010, 07:55 PM

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