How do you evaluate #∫sec^3(x) tan(x) dx#?

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Answer 1 · 2018-04-03T17:29:34

The answer is #=1/3sec^3x+C#

Perform this integral by substitution

Let #u=sec^3x#

#du=3sec^2xsecxtanxdx#

Therefore, the integral is

#I=intsec^3xtanxdx=1/3intdu#

#=1/3u#

#=1/3sec^3x+C#

Answer 2 · 2018-04-03T18:24:42

#int sec^3(x) tan(x) dx = 1/3 sec^3(x) + C#

The way I saw this was to notice that:

#sec^3(x) tan(x) = 1/cos^3(x) sin(x)/cos(x) = sin(x) cos^(-4)(x)#

and:

#d/(dx) cos^(-3)(x) = -3cos^(-4)(x)(-sin(x)) = 3 sin(x) cos^(-4)(x)#

So:

#int sec^3(x) tan(x) dx = 1/3 cos^(-3)(x) + C = 1/3 sec^3(x) + C#

Answer 3 · 2018-04-03T18:50:44

# 1/3sec^3x+C#.

#intsec^3xtanxdx=intsec^2x(secxtanx)dx#,

Since, #d/dx(secx)=secxtanx#, if we subst #secx=y#, we have,

#secxtanxdx=dy#.

#:. intsec^3xtanxdx=inty^2dy=1/3y^3#.

# rArr intsec^3xtanxdx=1/3sec^3x+C#.