Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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

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

Explanation:

`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`.