# 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
 P: 30 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|
 Sci Advisor HW Helper P: 11,835 $$\int \frac{\cos x}{\left( 1-\sin ^{2}x\right) ^{2}}dx=\allowbreak \int \frac{1}{\left( 1-t^{2}\right) ^{2}}\,dt$$ for the last integral, use simple fractions. $\sin x=t$ has been used.
P: 54

## integral of sec(x)^3

Ah, alright thanks. It was simpler than i thought it would be
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,713 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.

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