INTEGRATION 3

1/2[∫√tanx-√cotx dx +∫√tanx+√cotx dx]

give me the steps..........

first one is a tricky step ...then rewrite tan and cot in sin and cos

i m too lazy

otherwise
v = sqrt(tan(x))
dv = sec2(x) / (2 sqrt(tan(x)))
dv = 1 / (2 cos2(x) sqrt(tan(x)))

∫sqrt(tan(x)) dx
∫sqrt(tan(x)) dx/ (cos2(x) + sin2(x))
∫sqrt(tan(x)) dx/ (cos2(x) (1+tan2(x)))
∫tan(x) dx/ (cos2(x) sqrt(tan(x)) (1+tan2(x)))
∫2tan(x) dx/ (2 cos2(x) sqrt(tan(x)) (1+tan2(x)))
∫2v2 dv / (1+v4)

1+v4 = (v2-sqrt(2)v+1)(v2+sqrt(2)v+1)

So the same integral can now be expressed using this identity as

∫2v2 dv / ((v2-sqrt(2)v+1)(v2+sqrt(2)v+1))

And now, breaking it down into partial fractions, the integral becomes,

∫(sqrt(2)/2) v dv / (v2-sqrt(2)v+1) + (-sqrt(2)/2) v dv / (v2+sqrt(2)v+1)

Now this is starting to get tricky, so I'll break the first term into two integrals I1 and I2, and the second term into two more integrals I3 and I4.

∫sqrt(tan(x)) dx is I1 + I2 + I3 + I4, where

I1 = ∫(sqrt(2)/4)(2v-sqrt(2)) dv / (v2-sqrt(2)v+1)
I2 = ∫(1/2) dv / (v2-sqrt(2)v+1)
I3 = ∫(-sqrt(2)/4)(2v+sqrt(2)) dv / (v2+sqrt(2)v+1)
I4 = ∫(1/2) dv / (v2+sqrt(2)v+1)

Now I'll do each of I1, I2, I3, and I4 separately:

I1 = ∫(sqrt(2)/4)(2v-sqrt(2)) dv / (v2-sqrt(2)v+1)
This integral is of the form ∫du/u, which is ln|u|, so
I1 = (sqrt(2)/4) ln|v2-sqrt(2)v+1| + C1

I2 = ∫(1/2) dv / (v2-sqrt(2)v+1)
This can be converted into the form a/((av+b)2+1), if we let a=sqrt(2) and b=-1
I2 = ∫(sqrt(2)/2) sqrt(2)/(2v2-2sqrt(2)v+1+1)
I2 = ∫(sqrt(2)/2) sqrt(2)/((sqrt(2)v-1)2+1)
I2 = (sqrt(2)/2) atan(sqrt(2)v-1) + C2

I3 = ∫(-sqrt(2)/4)(2v+sqrt(2)) dv / (v2+sqrt(2)v+1)
Again, this is the ∫the ln form, so
I3 = (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + C3

I4 = ∫(1/2) dv / (v2+sqrt(2)v+1)
Again, this can be converted to the atan form, so
I4 = (sqrt(2)/2) atan(sqrt(2)v+1) + C4

To summarize,

I1 = (sqrt(2)/4) ln(v2-sqrt(2)v+1) + C1
I2 = (sqrt(2)/2) atan(sqrt(2)v-1) + C2
I3 = (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + C3
I4 = (sqrt(2)/2) atan(sqrt(2)v+1) + C4

The sum of which gives us the final answer,

∫sqrt(tan(x)) =
(sqrt(2)/4) ln(v2-sqrt(2)v+1) + (sqrt(2)/2) atan(sqrt(2)v-1) + (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + (sqrt(2)/2) atan(sqrt(2)v+1) + C =

(sqrt(2)/4) ln(tan(x)-sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))-1) +
(-sqrt(2)/4) ln(tan(x)+sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))+1) + C

copied from other source.....

It becomes

I₁= ∫(1+1/t²)dt/(t²+1/t²)

And now the master piece........

Let t-1/t = u

Then (1+1/t²)dt = du

t²+1/t² = u²+2 (verify yourselves)

Thus I₁=∫du/u²+2

= 1/√2 tan^-1(u/√2)

For I₂, divide both Nr and Dr by t² and let 1+1/t = v

Then dv=(1-1/t²) dt

Again, t²+1/t² = v²-2

I₂=∫dv/v²-2

=(1/2√2)logl(v-√2)/(v+√2)l

Combine I₁ and I₂ and substitute for u and v by t and again substitute for t as √tanx

HOPE U UNDERSTAND [1]

ya i understood guyzz so thanks ..........alot.............

my favourite method
I=1/2[∫√tanx-√cotx dx +∫√tanx+√cotx dx]

=1/2[∫√2{√sinx +√cosx}/√1-(sinx-cosx)² dx +∫1/2[√2∫{√sinx -√cosx}/√(sinx+cosx)²-1 dx]

=1/√2[∫dt/√1-t²-∫dm/√m²-1]

[taking sinx-cosx=p and sinx+cosx=m]

now it is simple.....

yeah :)

i too jus love that method ani :)