Evaluating the integral with trig

[m0gh]

Homework Statement

Evaluate the integral x^(-1/5) tan(x^(4/5)) dx

Homework Equations

The Attempt at a Solution

Set u = x^(4/5)
Therefore du = 4/5 x ^ (-1/5) dx

which gave me 5/4 ∫ tan(u) du

integral of tan is natural log of absolute value of sec(u)

sub in x^(4/5) for u and get 5/4 natural log of absolute value of sec(x^(4/5)) + c

I'm sorry if I didn't use the proper formatting as I'm not %100 how. I will attach an image to clarify my work. I used an indefinite integral calculator to check my work but it is giving a different answer. Can anyone verify that this solution is correct?

 

[scurty]

Homework Statement

Evaluate the integral x^(-1/5) tan(x^(4/5)) dx

Homework Equations

The Attempt at a Solution

Set u = x^(4/5)
Therefore du = 4/5 x ^ (-1/5) dx

which gave me 5/4 ∫ tan(u) du

integral of tan is natural log of absolute value of sec(u)

sub in x^(4/5) for u and get 5/4 natural log of absolute value of sec(x^(4/5)) + c

I'm sorry if I didn't use the proper formatting as I'm not %100 how. I will attach an image to clarify my work. I used an indefinite integral calculator to check my work but it is giving a different answer. Can anyone verify that this solution is correct?

##\ln{|\sec{x}|} = \ln{|\cos{x}^{-1}|} = -\ln{|\cos{x}|}##

Did thje online calculator have the answer in terms of cosine?

 

[m0gh]

The online calculator gives me this: (-5 Log[Cos[x^(4/5)]])/4 as the answer.

Try this link and let me know if it works. Never tried the clip'n share feature of wolframalpha before.

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427e7s6pgqqjsn

 

[m0gh]

Also can someone please explain why in class my teacher uses the natural log of the absolute value when integrating 1/x but when I use something like wolframalpha it always uses log. I have never had a teacher really explain the difference between these (I'm assuming they are interchangeable) and it would really help with my understanding. I've read online but I still don't truly understand what it means.

 

[scurty]

Also can someone please explain why in class my teacher uses the natural log of the absolute value when integrating 1/x but when I use something like wolframalpha it always uses log. I have never had a teacher really explain the difference between these (I'm assuming they are interchangeable) and it would really help with my understanding. I've read online but I still don't truly understand what it means.

Here is a better link: http://www.wolframalpha.com/input/?i=integral+of+x^(-1/5)+tan(x^(4/5))

Underneath the answer, it says that log(x) is the natural logarithm. if it was a different base, it wouldn't be written as ##\log_2{x}##, for example. http://www.wolframalpha.com/input/?i=log_2(x)

Right under the example, you can see the general formula for writing the log in any base in terms of the natural logarithm: ##\log_b{x}=\frac{\log{x}}{\log{b}}##. Note what this becomes when b = e.

The natural logarithm has a plethora of use in mathematics, more so than other bases, because of its relationship with e. In practice, it is shorter to write ln than log, so that may be why it is used.


To answer you other question one post up, they are equivalent. Read my other post to see why. When using online calculators, it is important to realize that there may be alternate forms to the answer.

Another hint would be to subtract your answer from the answer WolframAlpha gives to see if the result is 0. http://www.wolframalpha.com/input/?i=5/4*Log|Sec[x^(4/5)]|+5/4*+Log|Cos[x^(4/5)]|

 

[m0gh]

Okay, so now I understand that ln is just a logarithm with e as the base. What I don't get is the need for the absolute value bars that show up. Is this in the nature of a logarithm? (I don't understand logarithms very well at all.)

 

[LCKurtz]

Okay, so now I understand that ln is just a logarithm with e as the base. What I don't get is the need for the absolute value bars that show up. Is this in the nature of a logarithm? (I don't understand logarithms very well at all.)

You know the derivative of ##\ln x## is ##\frac 1 x##. The domain of ##\ln x## is ##x>0##. The domain of ##\frac 1 x## is ##x\ne 0##. So here the logarithm and its derivative seem to have different domains. If we use absolute values, then ##\ln|x|## has domain ##x\ne 0##, and it agrees with ##\ln x## when ##x>0##. What is the derivative of ##\ln|x|## if ##x<0##? Remembering that ##|x| = -x## if ##x<0##, we get$$
\frac d {dx}\ln |x| = \frac d {dx}\ln(-x) = \frac 1 {-x}\cdot (-1) = \frac 1 x$$by the chain rule if ##x<0##. So for all ##x\ne 0## the derivative of ##\ln |x|## is ##\frac 1 x##. The absolute value bars are often ignored, especially if the function given is ##\ln x## since its domain is automatically ##x>0##. Taking antiderivatives though, one should use the absolute values unless something in the problem specifies only positive ##x##.