Proofs of Selected Integrals

Solution, with a big hint by David Loeffler:

I will find òsqrt(tan(x)) using the substitution v = sqrt(tan(x))

First, a little differentiation of v:

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

Now, I will process the integral so that dv appears in it:

òsqrt(tan(x)) dx
òsqrt(tan(x)) dx/ (cos²(x) + sin²(x))
òsqrt(tan(x)) dx/ (cos²(x) (1+tan²(x)))
òtan(x) dx/ (cos²(x) sqrt(tan(x)) (1+tan²(x)))
ò2tan(x) dx/ (2 cos²(x) sqrt(tan(x)) (1+tan²(x)))
ò2v² dv / (1+v⁴)

David took us this far, and then gave one more hint, which led us to express 1+v⁴ this way:

1+v⁴ = (v²-sqrt(2)v+1)(v²+sqrt(2)v+1)

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

ò2v² dv / ((v²-sqrt(2)v+1)(v²+sqrt(2)v+1))

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

ò(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.

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

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

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

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

I₂ = ò(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 = ò(sqrt(2)/2) sqrt(2)/(2v²-2sqrt(2)v+1+1)
I₂ = ò(sqrt(2)/2) sqrt(2)/((sqrt(2)v-1)²+1)
I₂ = (sqrt(2)/2) atan(sqrt(2)v-1) + C₂

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

I₄ = ò(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,

òsqrt(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

Related Pages in this Website

Table of Integrals

Integration by Parts

Partial Fractions

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