# Integral of sec(x)^3

by imranq
Tags: integral, secx3
 P: 54 1. The problem statement, all variables and given/known data I'm just trying to solve this $$\int {\sec^3{x} dx$$ 2. Relevant equations $$sec^2{x} = 1 + tan^2{x}$$ 3. 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
 Sci Advisor HW Helper P: 11,948 $$\int \frac{\cos x}{\left( 1-\sin ^{2}x\right) ^{2}}dx=\allowbreak \int \frac{1}{\left( 1-t^{2}\right) ^{2}}\,dt$$ for the last integral, use simple fractions. $\sin x=t$ has been used.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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.