Antiderivative Definite Integrals

[cathy]

Homework Statement

So I did an entire antiderivative, and ended with this part:

sec(x)tan(x) + ln|sec(x) + tan(x)| + C

I have to do this with the lower bound of -pi/3 and 0.
When I do it, I should be getting 2√3 + ln(2+√3)

But, I'm getting (0+0)-(2*-√3 + ln(2-√3))
Which would give me 2√3-ln(2-√3)

Will someone tell me where I'm going wrong?
Does it have to do with the absolute value bars?

Thanks in advance.

 

[Saitama]

Homework Statement

So I did an entire antiderivative, and ended with this part:

sec(x)tan(x) + ln|sec(x) + tan(x)| + C

I have to do this with the lower bound of -pi/3 and 0.
When I do it, I should be getting 2√3 + ln(2+√3)

But, I'm getting (0+0)-(2*-√3 + ln(2-√3))
Which would give me 2√3-ln(2-√3)

Will someone tell me where I'm going wrong?
Does it have to do with the absolute value bars?

Thanks in advance.

Hi cathy!

Both answers are correct.

$$\ln(2-\sqrt{3})=\ln\left(\frac{1}{2+\sqrt{3}}\right)=-\ln(2+\sqrt{3})$$

 

[cathy]

Oh! Silly error on my end.
Thank you very much!

 

[cathy]

Wait, maybe another silly error on my end, but I'm having a similar problem.

The integral of secx dx with bounds of 0 and pi/4.
The antideriv of sec x is ln(secx+tanx)
so I plug everything in, and I am getting ln(sqrt2-1)
rather than the answer, which is ln(sqrt2+1)

Am I doing something wrong?

 

[Saitama]

Wait, maybe another silly error on my end, but I'm having a similar problem.

The integral of secx dx with bounds of 0 and pi/4.
The antideriv of sec x is ln(secx+tanx)

Yes. :)

so I plug everything in, and I am getting ln(sqrt2-1)

Not sure how you get this, what is the value of $\tan(\pi/4)$?

 

[cathy]

oh my, it is 1 and not 0.
Sorry!

Thank you so much, once again Pranav-Arora.
I'm just a bit tired :)
Thanks!

 

[Saitama]

oh my, it is 1 and not 0.
Sorry!

Thank you so much, once again Pranav-Arora.
I'm just a bit tired :)
Thanks!

Glad to help. :)