Can some one please verify integration of cscx = lntan(x/2) +c?

Question

and determine integration of x*csc(4-x^2)dx

Answer 1

∫ cscx dx = ∫ (1/sinx) dx =

let tan(x/2) = u →

(x/2) = arctan u →

x = 2arctan u →

dx = [2/(u² + 1)] du

now you have to express sinx in terms of tan(x/2), using double-angle identities:

sinx = 2 sin(x/2) cos(x/2) = [2 sin(x/2) cos(x/2)]/[sin²(x/2) + cos²(x/2)]

(since sin²(x/2) + cos²(x/2) = 1)

now divide both the numerator and the denominator by cos²(x/2) :

sinx = [2 sin(x/2) cos(x/2)/cos²(x/2)]/ {[sin²(x/2)/cos²(x/2)] + [cos²(x/2)/cos²(x/2)]} =

sinx = [2 sin(x/2) /cos(x/2)]/ {[sin²(x/2)/cos²(x/2)] + 1} = [2 tan(x/2) / [tan²(x/2) + 1]

thus, being tan(x/2) = u,

sinx = [2 tan(x/2) / [tan²(x/2) + 1] = [2u / (u² + 1)]

to sum up,

tan(x/2) = u

dx = [2/(u² + 1)] du

sinx = [2u / (u² + 1)]

then, substituting, you get:

∫ (1/sinx) dx = ∫ {1/ [2u / (u² + 1)]} [2/(u² + 1)] du = ∫ [(u² + 1)/ 2u] [2/(u² + 1)] du =

which simplifies into:

∫ (1/ u) du = ln | u | + C

thus, substituting back u = tan(x/2) you get:

∫ cscx dx =  ln | tan(x/2) | + C

∫ x csc(4 - x²) dx =

note that the integrand includes both the function (4 - x²) and something

(i.e. x) that can be easily turned into its derivative (that actually would be - 2x);

thus divide and multiply your integral by - 2:

(-1/2) ∫ (-2x) csc(4 - x²) dx =

(-1/2) ∫ csc(4 - x²) (-2x) dx =

let (4 - x²) = u

differentiate both sides:

d(4 - x²) = du →

- 2x dx = du

thus substitute, yielding:

(-1/2) ∫ csc(4 - x²) (-2x) dx = (-1/2) ∫ csc u du = (-1/2) ln | tan(u/2) | + C

(it has been verified previously)

finally, substituting back u = (4 - x²) you get:

∫ x csc(4 - x²) dx = (-1/2) ln | tan[(4 - x²)/2] | + C

I hope it helps..

Bye!

Answer 2

Since you are only asking for verification, we just have to differentiate ln tan(x/2) to see if we get csc x.

d/dx (ln tan(x/2))

= 1/tan(x/2) * sec^2 (x/2) * (1/2)

= cos(x/2) / sin(x/2) * 1/cos^2 (x/2) * 1/2

= 1/[2 sin(x/2) cos(x/2)]

= 1/sinx

= csc x

let u=4-x^2

du = -2xdx

Therefore the integral becomes

int (-1/2) csc u du

= -1/2 int csc u du

= -1/2 lntan(u/2) + c

= -1/2 lntan[(4-x^2)/2] + c

Answer 3

i got

ln{|sin x|/|cos x + 1|}

ANd

ii)

ln{|cos(x²-4}+1|/|sin(x²-4|]   /  2