# Integral of sec(x)^3

## 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

Last edited:

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

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

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

HallsofIvy
$$\int sec^3(x)dx= \int\frac{dx}{cos^3(x)}$$
$$\int \frac{cos(x)dx}{cos^4(x)}= \int\frac{cos(x)dx}{(1- sin^2(x))^2}$$
$$\int \frac{du}{(1-u^2)^2}= \int\frac{du}{(1-u)^2(1+u)^2}$$