Integral of sec^3(x) | Solving with Tan^2(x)

[imranq]

Homework Statement

I'm just trying to solve this

\int {\sec^3{x} dx

Homework Equations

sec^2{x} = 1 + tan^2{x}

The Attempt at a Solution

well i was able to simplify it to this:

\int {\sec{x}*\tan^2{x}} dx + \ln{|\sec{x} + \tan{x}|}

but I still was not able to find that new integral

 

[SunGod87]

INT[sec^3x]dx
= INT[secx.sec^2x]dx
By parts (differentiating secx and integrating sec^2x):
= INT[secx.sec^2x]dx
= secx.tanx - INT[secx.tan^2x]dx
= secx.tanx - INT[secx(sec^2x - 1)]dx
= secx.tanx - INT[sec^3x - secx]dx
So we have:

INT[sec^3x]dx = secx.tanx - INT[sec^3x - secx]dx
INT[sec^3x]dx = secx.tanx - INT[sec^3x]dx + INT[secx]dx
2 INT[sec^3x]dx = secx.tanx + INT[secx]dx
2 INT[sec^3x]dx = secx.tanx + Ln|secx + tanx|
INT[sec^3x]dx = 1/2 secx.tanx + 1/2 Ln|secx + tanx|

 

[dextercioby]

\int \frac{\cos x}{\left( 1-\sin ^{2}x\right) ^{2}}dx=\allowbreak \int \frac{1}{\left( 1-t^{2}\right) ^{2}}\,dt

for the last integral, use simple fractions. \sin x=t has been used.

 

[imranq]

Ah, alright thanks. It was simpler than i thought it would be

 

[HallsofIvy]

Another way: Since sec(x)= 1/cos(x),
\int sec^3(x)dx= \int\frac{dx}{cos^3(x)}
which is an odd power of cos(x). Multiply numerator and denominator by cos(x):
\int \frac{cos(x)dx}{cos^4(x)}= \int\frac{cos(x)dx}{(1- sin^2(x))^2}
Let u= sin(x) so du= cos(x)dx
\int \frac{du}{(1-u^2)^2}= \int\frac{du}{(1-u)^2(1+u)^2}
and, again, use partial fractions.