Question

Question #890e9

Answer 1

`int tanx/(sqrt(sec^2x-4)) dx = -1/2 arcsin(2cosx) + C`

Explanation:

Use the trigonometric identities:

`tanx =sinx/cosx`

`secx = 1/cosx`

so that:

`int tanx/(sqrt(sec^2x-4)) dx = int sinx/cosx 1/sqrt(1/cos^2x -4)dx`

`int tanx/(sqrt(sec^2x-4)) dx =int sinx/sqrt(1 -4cos^2x)dx`

substitute:

`t = 2cosx`

`dt = -2sinx`

`int tanx/(sqrt(sec^2x-4)) dx =-1/2 int dt/sqrt(1-t^2) = -1/2arcsint+C`

and undoing the substitution:

`int tanx/(sqrt(sec^2x-4)) dx = -1/2 arcsin(2cosx) + C`

Answer 2

`1/2sec^-1(sec(x)/2)+C`

Explanation:

We can also try letting `sec(x)=2sec(theta)`, implying that `sec(x)tan(x)dx=2sec(theta)tan(theta)d theta`. Then:

`I=inttan(x)/sqrt(sec^2(x)-4)dx=int(sec(x)tan(x)dx)/(sec(x)sqrt(sec^2(x)-4))`

`color(white)I=int(2sec(theta)tan(theta)d theta)/(2sec(theta)sqrt(4sec^2(theta)-4))=inttan(theta)/(2sqrt(sec^2(theta)-1))d theta`

And since `sec^2(theta)-1=tan^2(theta)`:

`I=1/2intd theta=1/2theta+C`

Reverse the substitution `sec(x)=2sec(theta)`:

`I=1/2sec^-1(sec(x)/2)+C`

Compared graphically, you can see that `1/2sec^-1(sec(x)/2)` and the other presented solution `-1/2sin^-1(2cos(x))` are different by only a constant, so they are both valid solutions.