# Integral of sec^3 (theta) d(theta)

• Mar 26th 2010, 08:09 AM
s3a
Integral of sec^3 (theta) d(theta)
How do I integrate sec^3 (theta) d(theta) ?

I am doing another problem and this is the step I am stuck on. I know all my previous steps are correct because math software told me but I'd like to be able to manually do this.

Any help would be greatly appreciated!
• Mar 26th 2010, 08:41 AM
Pulock2009
Quote:

Originally Posted by s3a
How do I integrate sec^3 (theta) d(theta) ?

I am doing another problem and this is the step I am stuck on. I know all my previous steps are correct because math software told me but I'd like to be able to manually do this.

Any help would be greatly appreciated!

1)express sec as cos
2)multiply numerator and denominator by cos(theta)
3)denominator becomes an even power of cos. so express it as sin
4)substitute sin(theta)=t
5)use partial fractions
6)integrate
• Mar 27th 2010, 06:44 AM
s3a
I subsitute for u instead of t because I am more familiar with it and I am at Integral of 1/(1-u^2)^2 du but am a bit lost now. Could you show me what to do with the partial fractions?

I don't know if the following is right:

1 = A(u+1)(u-1)^2 + B(u-1)^2 + C(u+1)^2 * (u-1) + D(u+1)^2
• Mar 27th 2010, 07:20 AM
skeeter
Quote:

Originally Posted by s3a
How do I integrate sec^3 (theta) d(theta) ?

I always thought the method of parts was a neat way to do it ...

$\displaystyle \int \sec{t} \cdot \sec^2{t} \, dt$

$\displaystyle u = \sec{t}$ ... $\displaystyle du = \sec{t}\tan{t} \, dt$

$\displaystyle dv = \sec^2{t} \, dt$ ... $\displaystyle v = \tan{t}$

$\displaystyle \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int \tan^2{t}\sec{t} \, dt$

$\displaystyle \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int (sec^2{t}-1)\sec{t} \, dt$

$\displaystyle \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int sec^3{t}-\sec{t} \, dt$

$\displaystyle \int \sec^3{t} \, dt = \sec{t}\tan{t} - \int sec^3{t} \, dt + \int \sec{t} \, dt$

$\displaystyle 2\int \sec^3{t} \, dt = \sec{t}\tan{t} + \int \sec{t} \, dt$

$\displaystyle 2\int \sec^3{t} \, dt = \sec{t}\tan{t} + \ln|\sec{t}+\tan{t}| + C$

$\displaystyle \int \sec^3{t} \, dt = \frac{1}{2}\left[\sec{t}\tan{t} + \ln|\sec{t}+\tan{t}|\right] + C$
• Mar 27th 2010, 10:06 AM
s3a
Just a question, what if I had a higher power though. Like what if I had Integral (sec^5(x))dx is there any easy method for that?