# Working with sec...any thoughts

• November 28th 2010, 02:37 PM
minusb
Working with sec...any thoughts
Hi,

I'm solving the integral of sec x.

I'm using a t sub ie $t = tan(x/2)$

I've integrated to $ln(|1+t|/|1-t|)$

Now I believe I have to multiply inside the brackets by $cos(x/2)+sin(x/2)$ so I'll end up with

$(cos^2(x/2)+sin^2(x/2)+2cos(x/2)sin(x/2))/cos(x)$

Could someone explain that to me?

Thank You.
• November 28th 2010, 02:46 PM
TheEmptySet
Quote:

Originally Posted by minusb
Hi,

I'm solving the integral of sec x.

I'm using a t sub ie $t = tan(x/2)$

I've integrated to $ln(|1+t|/|1-t|)$

Now I believe I have to multiply inside the brackets by $cos(x/2)+sin(x/2)$ so I'll end up with

$(cos^2(x/2)+sin^2(x/2)+2cos(x/2)sin(x/2))/cos(x)$

Could someone explain that to me?

Thank You.

try the substitution

$u=\sec(x)+\tan(x) \implies du = \sec(x)\tan(x)+\sec^2(x)\, dx = \sec(x) u \,dx$

This gives the integral

$\displaystyle \frac{1}{u}\,du=...$
• November 28th 2010, 02:57 PM
minusb
Thanks, I know that version... The question is asking to use t=tan(x/2) substitution.
• November 28th 2010, 03:07 PM
TheCoffeeMachine
Quote:

Originally Posted by minusb
I've integrated to $ln(|1+t|/|1-t|)$

Well, you have done it all! Just resubstitute $t$ and note that $1 = \tan\frac{\pi}{4}$, then use the tangent addition identity:

$\displaystyle \ln\left\{\frac{1+t}{1-t}\right\}+k = \ln\left\{\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right\}+k = \ln\left\{\frac{\tan{\frac{\pi}{4}}+\tan\frac{x}{2 }}{\tan{\frac{\pi}{4}}-\tan\frac{x}{2}}\right\}+k = \displaystyle \ln\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\righ t)\right\}+k.$
• November 29th 2010, 04:12 AM
minusb
Quote:

Originally Posted by TheCoffeeMachine

$\displaystyle \ln\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\righ t)\right\}+k.$

Is this a common identity that is equal to secx + tanx?
• November 29th 2010, 04:13 AM
minusb
Double post...
• November 29th 2010, 03:19 PM
TheCoffeeMachine
Quote:

Originally Posted by minusb
Is this a common identity that is equal to secx + tanx?

Why do you have to show that they are equal? This is the classical integral of $\displaystyle \sec{x}$!
If they have asked you to use that substitution, it should really be what they wanted!